# Find minimum number to be divided to make a number a perfect square

Given a positive integer n. Find the minimum number which divide n to make it a perfect square.

Examples:

```Input : n = 50
Output : 2
By Dividing n by 2, we get which is a perfect square.

Input : n = 6
Output : 6
By Dividing n by 6, we get which is a perfect square.

Input : n = 36
Output : 1
```

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

A number is perfect square if all prime factors appear even number of times. The idea is to find the prime factor of n and find each prime factor power. Now, find and multiply all the prime factor whose power is odd. The resultant of the multiplication is the answer.

## C++

 `// C++ program to find minimum number which divide n ` `// to make it a perfect square. ` `#include ` `using` `namespace` `std; ` ` `  `// Return the minimum number to be divided to make ` `// n a perfect square. ` `int` `findMinNumber(``int` `n) ` `{ ` `    ``int` `count = 0, ans = 1; ` ` `  `    ``// Since 2 is only even prime, compute its ` `    ``// power seprately. ` `    ``while` `(n%2 == 0) ` `    ``{ ` `        ``count++; ` `        ``n /= 2; ` `    ``} ` ` `  `    ``// If count is odd, it must be removed by dividing ` `    ``// n by prime number. ` `    ``if` `(count%2) ` `        ``ans *= 2; ` ` `  `    ``for` `(``int` `i = 3; i <= ``sqrt``(n); i += 2) ` `    ``{ ` `        ``count = 0; ` `        ``while` `(n%i == 0) ` `        ``{ ` `            ``count++; ` `            ``n /= i; ` `        ``} ` ` `  `        ``// If count is odd, it must be removed by ` `        ``// dividing n by prime number. ` `        ``if` `(count%2) ` `            ``ans *= i; ` `    ``} ` ` `  `    ``if` `(n > 2) ` `        ``ans *= n; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``int` `n = 72; ` `    ``cout << findMinNumber(n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find minimum number  ` `// which divide n to make it a perfect square. ` ` `  `class` `GFG ` `{ ` `    ``// Return the minimum number to be  ` `    ``// divided to make n a perfect square. ` `    ``static` `int` `findMinNumber(``int` `n) ` `    ``{ ` `        ``int` `count = ``0``, ans = ``1``; ` `     `  `        ``// Since 2 is only even prime,  ` `        ``// compute its power seprately. ` `        ``while` `(n % ``2` `== ``0``) ` `        ``{ ` `            ``count++; ` `            ``n /= ``2``; ` `        ``} ` `     `  `        ``// If count is odd, it must be removed by dividing ` `        ``// n by prime number. ` `        ``if` `(count % ``2` `== ``1``) ` `            ``ans *= ``2``; ` `     `  `        ``for` `(``int` `i = ``3``; i <= Math.sqrt(n); i += ``2``) ` `        ``{ ` `            ``count = ``0``; ` `            ``while` `(n % i == ``0``) ` `            ``{ ` `                ``count++; ` `                ``n /= i; ` `            ``} ` `     `  `            ``// If count is odd, it must be removed by ` `            ``// dividing n by prime number. ` `            ``if` `(count % ``2` `== ``1``) ` `                ``ans *= i; ` `        ``} ` `     `  `        ``if` `(n > ``2``) ` `            ``ans *= n; ` `     `  `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `n = ``72``; ` `        ``System.out.println(findMinNumber(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python program to find  ` `# minimum number which  ` `# divide n to make it a  ` `# perfect square. ` `import` `math ` ` `  `# Return the minimum  ` `# number to be divided  ` `# to make n a perfect  ` `# square. ` `def` `findMinNumber(n): ` `    ``count ``=` `0` `    ``ans ``=` `1` ` `  `    ``# Since 2 is only  ` `    ``# even prime, compute  ` `    ``# its power seprately. ` `    ``while` `n ``%` `2` `=``=` `0``: ` `        ``count ``+``=` `1` `        ``n ``/``/``=` `2` ` `  `    ``# If count is odd,  ` `    ``# it must be removed ` `    ``# by dividing n by  ` `    ``# prime number. ` `    ``if` `count ``%` `2` `is` `not` `0``: ` `        ``ans ``*``=` `2` ` `  `    ``for` `i ``in` `range``(``3``, (``int``)(math.sqrt(n)) ``+` `1``, ``2``): ` `        ``count ``=` `0` `        ``while` `n ``%` `i ``=``=` `0``: ` `            ``count ``+``=` `1` `            ``n ``/``/``=` `i ` ` `  `        ``# If count is odd, it  ` `        ``# must be removed by  ` `        ``# dividing n by prime  ` `        ``# number. ` `        ``if` `count ``%` `2` `is` `not` `0``: ` `            ``ans ``*``=` `i ` ` `  `    ``if` `n > ``2``: ` `        ``ans ``*``=` `n ` ` `  `    ``return` `ans ` ` `  `# Driver Code ` `n ``=` `72` `print``(findMinNumber(n)) ` ` `  `# This code is contributed ` `# by Sanjit_Prasad. `

## C#

 `// C# program to find minimum  ` `// number which divide n to ` `// make it a perfect square. ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Return the minimum number ` `    ``// to be divided to make  ` `    ``// n a perfect square. ` `    ``static` `int` `findMinNumber(``int` `n) ` `    ``{ ` `        ``int` `count = 0, ans = 1; ` `     `  `        ``// Since 2 is only even prime,  ` `        ``// compute its power seprately. ` `        ``while` `(n % 2 == 0) ` `        ``{ ` `            ``count++; ` `            ``n /= 2; ` `        ``} ` `     `  `        ``// If count is odd, it must  ` `        ``// be removed by dividing ` `        ``// n by prime number. ` `        ``if` `(count % 2 == 1) ` `            ``ans *= 2; ` `     `  `        ``for` `(``int` `i = 3; i <= Math.Sqrt(n);  ` `                                  ``i += 2) ` `        ``{ ` `            ``count = 0; ` `            ``while` `(n % i == 0) ` `            ``{ ` `                ``count++; ` `                ``n /= i; ` `            ``} ` `     `  `            ``// If count is odd, it must  ` `            ``// be removed by dividing n ` `            ``// by prime number. ` `            ``if` `(count % 2 == 1) ` `                ``ans *= i; ` `        ``} ` `     `  `        ``if` `(n > 2) ` `            ``ans *= n; ` `     `  `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `n = 72; ` `        ``Console.WriteLine(findMinNumber(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` 2) ` `        ``\$ans` `*= ``\$n``; ` ` `  `    ``return` `\$ans``; ` `} ` ` `  `    ``// Driver Code ` `    ``\$n` `= 72; ` `    ``echo` `findMinNumber(``\$n``), ``"\n"``; ` `     `  `// This code is contributed by ajit. ` `?> `

Output:

```2
```

This article is contributed by Anuj Chauhan. 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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Improved By : vt_m, jit_t, Sanjit_Prasad

Article Tags :
Practice Tags :

Be the First to upvote.

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