# 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

## CPP

 `// C++ program to find product of ` `// unique prime factors of a number ` `#include ` `using` `namespace` `std; ` ` `  `long` `long` `int` `productPrimeFactors(``int` `n) ` `{ ` `    ``long` `long` `int` `product = 1; ` ` `  `    ``for` `(``int` `i = 2; i <= n; i++) { ` ` `  `        ``// Checking if 'i' is factor of num ` `        ``if` `(n % i == 0) { ` ` `  `            ``// Checking if 'i' is a Prime number ` `            ``bool` `isPrime = ``true``; ` `            ``for` `(``int` `j = 2; j <= i / 2; j++) { ` `                ``if` `(i % j == 0) { ` `                    ``isPrime = ``false``; ` `                    ``break``; ` `                ``} ` `            ``} ` ` `  `            ``// condition if 'i' is Prime number ` `            ``// as well as factor of num ` `            ``if` `(isPrime) { ` `                ``product = product * i; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `product; ` `} ` ` `  `// driver function ` `int` `main() ` `{ ` `    ``int` `n = 44; ` `    ``cout << productPrimeFactors(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find product of ` `// unique prime factors of a number. ` ` `  `class` `GFG { ` `    ``public` `static` `long` `productPrimeFactors(``int` `n) ` `    ``{ ` `        ``long` `product = ``1``; ` ` `  `        ``for` `(``int` `i = ``2``; i <= n; i++) { ` `            ``// Checking if 'i' is factor of num ` `            ``if` `(n % i == ``0``) { ` ` `  `                ``// Checking if 'i' is a Prime number ` `                ``boolean` `isPrime = ``true``; ` `                ``for` `(``int` `j = ``2``; j <= i / ``2``; j++) { ` `                    ``if` `(i % j == ``0``) { ` `                        ``isPrime = ``false``; ` `                        ``break``; ` `                    ``} ` `                ``} ` ` `  `                ``// condition if 'i' is Prime number ` `                ``// as well as factor of num ` `                ``if` `(isPrime) { ` `                    ``product = product * i; ` `                ``} ` `            ``} ` `        ``} ` `        ``return` `product; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``44``; ` `        ``System.out.print(productPrimeFactors(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by _omg `

## 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 `

## C#

 `// C# program to find product of ` `// unique prime factors of a number. ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find product of unique ` `    ``// prime factors of a number ` `    ``public` `static` `long` `productPrimeFactors(``int` `n) ` `    ``{ ` `        ``long` `product = 1; ` ` `  `        ``for` `(``int` `i = 2; i <= n; i++) { ` ` `  `            ``// Checking if 'i' is factor of num ` `            ``if` `(n % i == 0) { ` ` `  `                ``// Checking if 'i' is a Prime number ` `                ``bool` `isPrime = ``true``; ` `                ``for` `(``int` `j = 2; j <= i / 2; j++) { ` `                    ``if` `(i % j == 0) { ` `                        ``isPrime = ``false``; ` `                        ``break``; ` `                    ``} ` `                ``} ` ` `  `                ``// condition if 'i' is Prime number ` `                ``// as well as factor of num ` `                ``if` `(isPrime) { ` `                    ``product = product * i; ` `                ``} ` `            ``} ` `        ``} ` `        ``return` `product; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 44; ` `        ``Console.Write(productPrimeFactors(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal `

## PHP

 ` `

Output :

```22
```

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

## CPP

 `// C++ program to find product of ` `// unique prime factors of a number ` `#include ` `using` `namespace` `std; ` ` `  `// A function to print all prime ` `// factors of a given number n ` `long` `long` `int` `productPrimeFactors(``int` `n) ` `{ ` `    ``long` `long` `int` `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` `(``int` `i = 3; i <= ``sqrt``(n); i = i + 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; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 44; ` `    ``cout << productPrimeFactors(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find product of ` `// unique prime factors of a number. ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG { ` `    ``public` `static` `long` `productPrimeFactors(``int` `n) ` `    ``{ ` `        ``long` `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` `(``int` `i = ``3``; i <= Math.sqrt(n); i = i + ``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; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``44``; ` `        ``System.out.print(productPrimeFactors(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by _omg `

## 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 `

## C#

 `// C# program to find product ` `// of unique prime factors ` `// of a number. ` `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``// Function to find product ` `    ``// of prime factors ` `    ``public` `static` `long` `productPrimeFactors(``int` `n) ` `    ``{ ` `        ``long` `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` `(``int` `i = 3; i <= Math.Sqrt(n); ` `             ``i = i + 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; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `n = 44; ` `        ``Console.Write(productPrimeFactors(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by parashar... `

## PHP

 ` 2) ` `        ``\$product` `= ``\$product` `* ``\$n``; ` ` `  `    ``return` `\$product``; ` `} ` ` `  `// Driver Code ` `\$n` `= 44;  ` `echo` `productPrimeFactors(``\$n``);  ` ` `  `// This code is contributed by ajit ` `?> `

Output :

```22
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : parashar, nitin mittal, jit_t

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.