### Python Program to Find Prime Factors of a Number

In this post I am going to write a program in python that finds all the prime factors of a number. I am going to start by writing an empty function and a test.

def get_prime_factors(number): prime_factors = [] return prime_factors if __name__ == "__main__": n = 8 expected = [2, 2, 2] result = get_prime_factors(n) assert expected == result, resultNow, if you run the program above, it will give an AssertionError, as we are returning an empty list. Let's write some code to make our test pass.

def get_prime_factors(number): prime_factors = [] while number % 2 == 0: prime_factors.append(2) number = number // 2 return prime_factorsThe program will work for multiple of 2's. Our next task is to find the other prime factors. For example, prime factors of 10 are 2 and 5. We shall add this test case first and then write code to pass this test. If you have studied prime numbers before, then you might remember that if the number itself is not prime, then you will find all the prime factors in the range from 2 to square root of n. So we shall use a loop to try every number upto square root of n to check if that number is a factor of n. And as we already tried to divide the number by 2, we can try only odd numbers.

from math import sqrt def get_prime_factors(number): prime_factors = [] while number % 2 == 0: prime_factors.append(2) number = number // 2 sq_root = int(sqrt(number)) for m in range(3, sq_root+1, 2): if number % m == 0: prime_factors.append(m) number = number // m return prime_factorsWe made some good progress. However, the get_prime_factors function above won't pass the test case for 10. Because in the first while loop, number is divisible by 2, so we shall add 2 to the list of prime factors. Then we divide the number by 2, so it becomes 5 (10 // 2). And the value of sq_root is 2, so the value of m in the for loop will be 3 (for m in range(3, 3, 2)). So we didn't get anything out of that loop. Because here 5 itself is a prime number. So after the for loop, we need to check if number is still greater than 1. And if it is, then we can safely assume that it's a prime number (and a prime factor of number), thus add to our list of prime factors. If you spend some time, you will be able to complete the code by yourself. And don't forget to test the code with different values of n (1, 2, 4, 5, 25, 100, 101). I hope you found it useful.

You can find the full code here.

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