Python Program for Product of Unique Prime Factors of a Number

Given a number n, the task is to find the product of all its unique prime factors. A prime factor is a prime number that divides n exactly without leaving a remainder. For Examples:

Input: 10
Output: Product is 10
Explanation: prime factors of 10 are 2 and 5 and their product is 10.

Let's explore different methods to find the product of unique prime factors of a number in python.

Using SymPy Library

In this method, we use the built-in primefactors() function that directly returns a list of unique prime factors of a number. We then compute their product using math.prod().

from sympy import primefactors
import math

n = 44
f = primefactors(n)
p = math.prod(f)
print(p)

Output

22

Using NumPy Product Function

This approach is similar to the SymPy method but uses numpy.prod() for computing the product efficiently when handling large factor lists.

from sympy import primefactors
import numpy as np

n = 44
f = primefactors(n)
p = np.prod(f)
print(p)

Output

22

Using Math Module

In this method, we divide the number by 2 until it’s no longer divisible, then check divisibility by odd numbers up to its square root. Each prime divisor is multiplied only once to form the final product.

import math
n = 44
p = 1

if n % 2 == 0:
    p *= 2
    while n % 2 == 0:
        n //= 2

for i in range(3, int(math.sqrt(n)) + 1, 2):
    if n % i == 0:
        p *= i
        while n % i == 0:
            n //= i

if n > 2:
    p *= n
print(p)

22

Explanation:

• Uses math.sqrt() to loop only up to the square root of n for efficiency and multiplies 2 once if it divides n.
• Checks odd divisors from 3 onwards and multiplies each distinct prime factor once.
• If a number greater than 2 remains, it’s a prime factor and multiplied at the end.

Using Iterative Division

In this method, we check each number from 2 to n. If it divides n, we check whether it’s prime by testing divisibility up to its half. Each prime divisor contributes once to the final product.

n = 44
p = 1

for i in range(2, n + 1):
    if n % i == 0:
        res = True
        for j in range(2, int(i / 2) + 1):
            if i % j == 0:
                res = False
                break
        if res:
            p *= i

print(p)

22

Explanation:

• Iterates through all possible factors of n.
• For each factor, checks if it’s prime by trying to divide it by smaller numbers.
• Multiplies the factor with product only if it’s prime.