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Ý tưởng là xem phần còn lại của mọi phần tử khi chia cho 3. Một tập hợp các phần tử chỉ có thể tạo thành một nhóm nếu tổng các phần tử còn lại của chúng là bội số của 3. Show Ví dụ. 8, 4, 12. Bây giờ, phần còn lại lần lượt là 2, 1 và 0. Điều này có nghĩa là 8 cách bội số 3 giây (6) 2 khoảng cách, 4 cách bội số 3 giây (3) 1 khoảng cách và 12 cách bội số 0. Vì vậy, chúng ta có thể viết tổng dưới dạng 8 (có thể viết là 6+2), 4 (có thể viết là 3+1) và 12 (có thể viết là 12+0). Bây giờ tổng của 8, 4 và 12 có thể được viết là 6+2+3+1+12+0. Bây giờ, 6+3+12 sẽ luôn chia hết cho 3 vì tất cả các số hạng đều là bội của 3. Bây giờ, chúng ta chỉ cần kiểm tra xem 2+1+0 (số dư) có chia hết cho 3 hay không để tổng có chia hết cho 3. 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result. Thực hiện C++1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.79 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.80 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.81
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.82 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.83 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.84 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.85
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.0 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.1 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.3 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.5 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.10 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.12 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.14 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.16 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.19
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.792 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.793
_______09____3795 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.798 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.800
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.802 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.804 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.806
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.808 _______09____3810 _______09____3812 _______09____3814
_______09____3816 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.818 _______09____3820
_______09____3822 _______09____3824 _______09____3826
_______09____3828 _______09____3830 _______09____3832
_______09____3828 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.836 _______09____3838
_______09____3840 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.843 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.844
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.845 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.847 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.851 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.854 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.855 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.856 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.855 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.858 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.00 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.01 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.02 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.03 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.06 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.844
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.08 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.09 C1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.10 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.80 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.81
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.13
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.14 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.15 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.3 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.5 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.24 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.26 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.19
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.792 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.793
_______09____3795 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.798 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.800
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.42 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.806
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.46 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.48 _______09____3814
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.52 _______09____3820
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.56 _______09____3826
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.60 _______09____3832
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.64 _______09____3838
_______09____3840 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.843 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.844
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.73 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.847 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.851 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.854 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.855 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.856 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.855 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.858 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.88 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.89 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.90 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.91 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.06 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.844 Java1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.96 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.80 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.81 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.99 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.100 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.14 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.15
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.3 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.5 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.24 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.26 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.121 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.122 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.124 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.130 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.133
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.136 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.138 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.793
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.795 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.144 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.146 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.147 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.148 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.149 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.150
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.42 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.154 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.156 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.158 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.160 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.162
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.46 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.48 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.168 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.170 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.172
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.52 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.176 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.156 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.158 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.182 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.158 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.186 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.187 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.138
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.56 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.176 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.156 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.158 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.182 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.158 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.186 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.187 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.138
_______3799____060 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.154 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.156 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.158 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.182 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.158 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.186 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.187 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.162
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.64 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.168 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.170 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.170 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.172
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.840 ________ 3799 ________ 3842 ________ 3843 _______09____3844
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7939 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7940 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7941 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7942 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7946 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.122 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7948 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7951 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.149 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.187 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7956 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7960 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.130 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7964 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7966____001 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7968 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7969 _______09____3844 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.844 Python31. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7973 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7974 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7975
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7976 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7977 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7978 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7979 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7980
_______09____37982 _______09____37984 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7986 _______09____37988 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7990 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7992______37993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7994 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8002 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8004 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9____38008 _______09____38010 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8013 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8014 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8015 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.89 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8018 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8020 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8021 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.149 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000______38024 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159
_______09____38028 _______09____38030 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8004 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8024 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8035 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.160 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8046
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9____38048 _______09____38050 _______09____38052 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8004 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8024 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8057______1159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8057 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000
_______09____38065 _______09____38067 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8004 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8024 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8046 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000__ 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.178 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.178 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.178
_______09____38092 _______09____38094 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8004 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8024 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039_______1159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8046 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000__ 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.178 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.178 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.178
_______09____38119 _______09____38121 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8004 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8024 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8035 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8046 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8039 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8088 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8089 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.187 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8046
_______09____38119 _______09____38149 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8004 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8024 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8057______1125 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8057 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.159 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8057 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000
_______09____38166 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8004
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8170 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7994 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.149 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126______1187 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7956 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.171 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7960 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8000 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8184 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8186 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8187
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8188 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.89 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8190 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8192____02____38194
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8195 C#1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8196 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.80 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.81 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.83 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8200
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.99 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.100 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 _______09____38206 _______09____38208 _______09____38210
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.3 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8215 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8222 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799____38224 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8226 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799____38228 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8231 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.122 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8234 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.133
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.793 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.792
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799____38244 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8246 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8249 _______1147____38251
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799____38253 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8255 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8257
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8259 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8261 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8263 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.814
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.816 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.818 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799____38271 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8192____38273
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.822 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.824 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8279 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8192 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8281
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.828 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.830 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8287 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7968____38289
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.828 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.836 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8295
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.840 ________ 3799 ________ 3842 ________ 3843 _______09____3844
_______09____3845 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7939 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7940 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7941 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8309 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7946 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.122 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7948 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8318 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8321 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.799 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8323 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.01 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8325____37969 _______09____3844 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.844
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8330 PHP1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8331 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8332 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8333
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.0 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.1 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8336 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.3____38338 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.126 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8340 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8046 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.10 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.12 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.14 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.16 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8352 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.7993____38354 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8355
_______09____3793 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8359 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8360
_______09____38362 _______09____38364 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.9 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.797______089 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8368 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8360 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8368 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8371 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8340 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.138 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8368 1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.8375 |