What is the least number which when divided by 4 5 6 and 7 leaves?

The least number which when divided by 4, 5 and 6 leaves remainder 1, 2 and 3 respectively, is :
[A]57
[B]59
[C]61
[D]63

57
Here 4 – 1 = 3, 5 – 2 = 3, 6 – 3 = 3
∴ The required Number = LCM of [4, 5, 6] – 3
= 60 – 3 = 57.
Hence option [A] is correct answer.

• Remainders

The smallest number which when divided by 4, 6 or 7 leaves a remainder of 2, is

1. 86
2. 80
3. 62
4. 44

Answer

Smallest number will be LCM [4,6,7] + 2.

LCM of 4, 6 and 7 is 84.

So, the smallest number leaving remainder 2 will be 86.

The correct option is A.

Solution

Step1:

Lets evaluate the LCM of 2,3,4,5 and 6.

22, 3,4,5, 621, 3,2,5, 331, 3,1,5, 351, 1,1,5, 11, 1,1,1, 1

⇒ LCM of 2,3,4,5 and 6

=2×2×3×5=60

LCM of 2,3,4,5 and 6 is 60.

Step2:

Since, the required number leaves the remainder 1 when it is divisible by 2,3,4,5,6 and no remainder when it is divisible by 7.

Therefore the required number is of the form 60x+1

⇒60x+1 is a multiple of 7

Now, lets check for 60x+1 is divisible by 7 by substituting the natural numbers sequentially,

60[1]+1=61=[7×8]+3 is not divisible by 7.

60[2]+1=121=[7×17]+2 is not divisible by 7.

60[3]+1=181=[7×25]+6 is not divisible by 7.

60[4]+1=241=[7×34]+3 is not divisible by 7.

60[5]+1=301=[7×43]+0 is divisible by 7.

So, the least number is 301 when divided by 2,3,4,5,6 leaves a remainder 1, but when divided by 7, there will be no remainder.

What is the least number which when divided by 4 5 6 and 7 leaves a remainder 3 but when divided by 9 leaves no remainder?

What is the smallest number which when divided by each of 4/5 and 7 leaves a remainder of 2?

What is the smallest multiple of 13 which if divided by 4 5 6 7 leaves remainder 3 each time?

What is the smallest number which when divided by 3 4 5 6 7 leaves a remainder 2 in each case and is divisible by 11?