# Minimum divisor of a number to make the number perfect cube

• Last Updated : 08 Apr, 2021

Given a positive integer N, the task is to find the minimum divisor by which it shall be divided to make it a perfect cube. If N is already a perfect cube, then print 1.
Examples:

```Input : N = 128
Output : 2
By Dividing N by 2, we get 64 which is a perfect cube.

Input : n = 6
Output : 6
By Dividing N by 6, we get 1 which is a perfect cube.

Input : n = 64
Output : 1```

Any number is a perfect cube if all prime factors of it appear in multiples of 3, as you can see in the below figure.

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Therefore, the idea is to find the prime factorization of N and find power of each prime factor. Now, find and multiply all the prime factors whose power is not divisible by 3 as primeFactor*power%3. The resultant of the multiplication is the answer.
Below is the implementation of the above approach:

## C++

 `// C++ program to find minimum number which divide n``// to make it a perfect cube``#include ``using` `namespace` `std;` `// Returns the minimum divisor``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 not divisible by 3,``    ``// it must be removed by dividing``    ``// n by prime number power.``    ``if` `(count % 3 != 0)``        ``ans *= ``pow``(2, (count % 3));` `    ``for` `(``int` `i = 3; i <= ``sqrt``(n); i += 2) {``        ``count = 0;``        ``while` `(n % i == 0) {``            ``count++;``            ``n /= i;``        ``}` `        ``// If count is not divisible by 3,``        ``// it must be removed by dividing``        ``// n by prime number power.``        ``if` `(count % 3 != 0)``            ``ans *= ``pow``(i, (count % 3));``    ``}` `    ``// if n is a prime number``    ``if` `(n > 2)``        ``ans *= n;` `    ``return` `ans;``}` `// Driven Program``int` `main()``{``    ``int` `n = 128;``    ``cout << findMinNumber(n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find minimum number which divide n``// to make it a perfect cube``class` `GFG{`` ` `// Returns the minimum divisor``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 not divisible by 3,``    ``// it must be removed by dividing``    ``// n by prime number power.``    ``if` `(count % ``3` `!= ``0``)``        ``ans *= Math.pow(``2``, (count % ``3``));`` ` `    ``for` `(``int` `i = ``3``; i <= Math.sqrt(n); i += ``2``) {``        ``count = ``0``;``        ``while` `(n % i == ``0``) {``            ``count++;``            ``n /= i;``        ``}`` ` `        ``// If count is not divisible by 3,``        ``// it must be removed by dividing``        ``// n by prime number power.``        ``if` `(count % ``3` `!= ``0``)``            ``ans *= Math.pow(i, (count % ``3``));``    ``}`` ` `    ``// if n is a prime number``    ``if` `(n > ``2``)``        ``ans *= n;`` ` `    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``128``;``    ``System.out.print(findMinNumber(n) +``"\n"``);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to find minimum number which divide n``# to make it a perfect cube` `# Returns the minimum divisor``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 not divisible by 3,``    ``# it must be removed by dividing``    ``# n by prime number power.``    ``if` `(count ``%` `3` `!``=` `0``):``        ``ans ``*``=` `pow``(``2``, (count ``%` `3``));` `    ``for` `i ``in` `range``(``3``, ``int``(``pow``(n, ``1``/``2``)), ``2``):``        ``count ``=` `0``;``        ``while` `(n ``%` `i ``=``=` `0``):``            ``count ``+``=` `1``;``            ``n ``/``=` `i;``        ` `        ``# If count is not divisible by 3,``        ``# it must be removed by dividing``        ``# n by prime number power.``        ``if` `(count ``%` `3` `!``=` `0``):``            ``ans ``*``=` `pow``(i, (count ``%` `3``));``    ` `    ``# if n is a prime number``    ``if` `(n > ``2``):``        ``ans ``*``=` `n;` `    ``return` `ans;` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `128``;``    ``print``(findMinNumber(n));` `# This code is contributed by 29AjayKumar`

## C#

 `// C# program to find minimum number which divide n``// to make it a perfect cube``using` `System;` `class` `GFG{``  ` `// Returns the minimum divisor``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 not divisible by 3,``    ``// it must be removed by dividing``    ``// n by prime number power.``    ``if` `(count % 3 != 0)``        ``ans *= (``int``)Math.Pow(2, (count % 3));``  ` `    ``for` `(``int` `i = 3; i <= Math.Sqrt(n); i += 2) {``        ``count = 0;``        ``while` `(n % i == 0) {``            ``count++;``            ``n /= i;``        ``}``  ` `        ``// If count is not divisible by 3,``        ``// it must be removed by dividing``        ``// n by prime number power.``        ``if` `(count % 3 != 0)``            ``ans *= (``int``)Math.Pow(i, (count % 3));``    ``}``  ` `    ``// if n is a prime number``    ``if` `(n > 2)``        ``ans *= n;``  ` `    ``return` `ans;``}``  ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 128;``    ``Console.Write(findMinNumber(n) +``"\n"``);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`2`

Time Complexity: O(sqrt(n))

Auxiliary Space: O(1)

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