How to find the factors of 320 in python?

Source Code

# Python Program to find the factors of a number

# This function computes the factor of the argument passed
def print_factors[x]:
   print["The factors of",x,"are:"]
   for i in range[1, x + 1]:
       if x % i == 0:
           print[i]

num = 320

print_factors[num]

Output

The factors of 320 are:
1
2
4
5
8
10
16
20
32
40
64
80
160
320

Note: To find the factors of another number, change the value of num.

In this program, the number whose factor is to be found is stored in num, which is passed to the print_factors[] function. This value is assigned to the variable x in print_factors[].

In the function, we use the for loop to iterate from i equal to x. If x is perfectly divisible by i, it's a factor of x.

FAQs on Factors of 320

What are Factors of 320?

The factors of 320 are 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160 and 320

Is 320 a Perfect Square?

No, 320 is not a perfect square, because it cannot be factored as number × number.

What are the Factors of 320 and 350?

• The factors of 320 are 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160 and 320
• The factors of 350 are 1, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175 and 350

What are Prime Factors of 320?

The prime factors of 320 are 2 and 5.

What is the Sum of the Factors of 320?

The factors of 320 are 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160 and 320. The sum of the factors of 320 = 762.

Here is an example if you want to use the primes number to go a lot faster. These lists are easy to find on the internet. I added comments in the code.

# //primes.utm.edu/lists/small/10000.txt
# First 10000 primes

_PRIMES = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 
        31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 
        73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 
        127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 
        179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 
        233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 
        283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 
        353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 
        419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 
        467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 
        547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 
        607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 
        661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 
        739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 
        811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 
        877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 
        947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 
# Mising a lot of primes for the purpose of the example
]


from bisect import bisect_left as _bisect_left
from math import sqrt as _sqrt


def get_factors[n]:
    assert isinstance[n, int], "n must be an integer."
    assert n > 0, "n must be greather than zero."
    limit = pow[_PRIMES[-1], 2]
    assert n