Python Program for Product of unique prime factors of a number
Last Updated :
23 Jul, 2025
Given a number n, we need to find the product of all of its unique prime factors. Prime factors: It is basically a factor of the number that is a prime number itself. Examples:
Input: num = 10
Output: Product is 10
Explanation:
Here, the input number is 10 having only 2 prime factors and they are 5 and 2.
And hence their product is 10.
Input : num = 25
Output: Product is 5
Explanation:
Here, for the input to be 25 we have only one unique prime factor i.e 5.
And hence the required product is 5.
Method 1 (Simple) Using a loop from i = 2 to n and check if i is a factor of n then check if i is prime number itself if yes then store product in product variable and continue this process till i = n.
Python3
# Python program to find sum of given
# series.
def productPrimeFactors(n):
product = 1
for i in range(2, n+1):
if (n % i == 0):
isPrime = 1
for j in range(2, int(i/2 + 1)):
if (i % j == 0):
isPrime = 0
break
# condition if \'i\' is Prime number
# as well as factor of num
if (isPrime):
product = product * i
return product
# main()
n = 44
print(productPrimeFactors(n))
# Contributed by _omg
Output:
22
Time complexity: O(n^2/2)
Auxiliary space: O(1)
Method 2 (Efficient) : The idea is based on Efficient program to print all prime factors of a given number
Python3
# Python program to find product of
# unique prime factors of a number
import math
def productPrimeFactors(n):
product = 1
# Handle prime factor 2 explicitly so that
# can optimally handle other prime factors.
if (n % 2 == 0):
product *= 2
while (n%2 == 0):
n = n/2
# n must be odd at this point. So we can
# skip one element (Note i = i +2)
for i in range (3, int(math.sqrt(n)), 2):
# While i divides n, print i and
# divide n
if (n % i == 0):
product = product * i
while (n%i == 0):
n = n/i
# This condition is to handle the case when n
# is a prime number greater than 2
if (n > 2):
product = product * n
return product
# main()
n = 44
print (int(productPrimeFactors(n)))
# Contributed by _omg
Output:
22
Time complexity: O(sqrt(n)), where n is the input number.
Auxiliary space: O(1), as the program only uses a constant amount of memory to store the product and the loop variables.
Please refer complete article on Product of unique prime factors of a number for more details!
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