How do you find the factor of a number 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.

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.

# http://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 <= limit, "n is greather then the limit of {0}".format(limit)
    result = set((1, n))
    root = int(_sqrt(n))
    primes = [t for t in get_primes_smaller_than(root + 1) if not n % t]
    result.update(primes)  # Add all the primes factors less or equal to root square
    for t in primes:
        result.update(get_factors(n/t))  # Add all the factors associted for the primes by using the same process
    return sorted(result)


def get_primes_smaller_than(n):
    return _PRIMES[:_bisect_left(_PRIMES, n)]

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In order to find factors of a number, we have to run a loop over all numbers from 1 to itself and see if it is divisible.

Example

num=int(input("enter a number"))
factors=[]
for i in range(1,num+1):
    if num%i==0:
       factors.append(i)

print ("Factors of {} = {}".format(num,factors))

If i is able to divide num completely, it is added in the list. Finally the list is diaplayed as the factors of given number

Output

enter a number75
Factors of 75 = [3, 5, 15, 25, 75]

Jayashree

Updated on 21-Feb-2020 12:59:32

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