Python Program for Product of unique prime factors of a number

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.

# 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

Method 2 (Efficient) :
The idea is based on Efficient program to print all prime factors of a given number

# 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

Output:

22

Please refer complete article on Product of unique prime factors of a number for more details!