# Minimize the value of N by applying the given operations

Given an integer *N*, below operations can be performed any number of times on *N*:

- Multiply
*N*by any positive integer*X*i.e.*N = N * X*. - Replace
*N*with square root of*N*(*N*must be an integer) i.e.*N = sqrt(N)*.

The task is to find the minimum integer to which *N* can be reduced with the above operations.

**Examples:**

Input:N = 20

Output:10

We can multiply 20 by 5, then take sqrt(20*5) = 10, this is the minimum number that 20 can be reduced to with the given operations.

Input: N = 36

Output:6

Take sqrt(36). Number 6 can’t be reduced further.

**Approach:**

- First factorize the number
*N*. - Say,
*12*has factors*2, 2 and 5*. Only the factors that are repeating can be reduced with*sqrt(n)*i.e.*sqrt(2*2) = 2*. - The numbers appearing only once in the factors cannot be further reduced.
- So, the final answer will be the product of all the distinct prime factors of number
*N*

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `#define ll long long int ` `using` `namespace` `std; ` ` ` `// function to return the product of ` `// distinct prime factors of a number ` `ll minimum(ll n) ` `{ ` ` ` `ll product = 1; ` ` ` ` ` `// find distinct prime ` ` ` `for` `(` `int` `i = 2; i * i <= n; i++) { ` ` ` `if` `(n % i == 0) { ` ` ` `while` `(n % i == 0) ` ` ` `n = n / i; ` ` ` `product = product * i; ` ` ` `} ` ` ` `} ` ` ` `if` `(n >= 2) ` ` ` `product = product * n; ` ` ` ` ` `return` `product; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `ll n = 20; ` ` ` `cout << minimum(n) << endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the above approach ` `import` `java.util.*; ` ` ` `class` `solution ` `{ ` ` ` ` ` `// function to return the product of ` ` ` `// distinct prime factors of a number ` `static` `int` `minimum(` `int` `n) ` `{ ` ` ` `int` `product = ` `1` `; ` ` ` ` ` `// find distinct prime ` ` ` `for` `(` `int` `i = ` `2` `; i * i <= n; i++) { ` ` ` `if` `(n % i == ` `0` `) { ` ` ` `while` `(n % i == ` `0` `) ` ` ` `n = n / i; ` ` ` `product = product * i; ` ` ` `} ` ` ` `} ` ` ` `if` `(n >= ` `2` `) ` ` ` `product = product * n; ` ` ` ` ` `return` `product; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String arr[]) ` `{ ` ` ` `int` `n = ` `20` `; ` ` ` `System.out.println(minimum(n)); ` ` ` `} ` `} ` `//This code is contributed by ` `//Surendra_Gangwar ` |

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## Python3

`# Python3 implementation of above approach ` ` ` `# function to return the product ` `# of distinct prime factors of a ` `# numberdef minSteps(str): ` `def` `minimum(n): ` ` ` ` ` `product ` `=` `1` ` ` ` ` `# find distinct prime ` ` ` `i ` `=` `2` ` ` `while` `i ` `*` `i <` `=` `n: ` ` ` `if` `n ` `%` `i ` `=` `=` `0` `: ` ` ` `while` `n ` `%` `i ` `=` `=` `0` `: ` ` ` `n ` `=` `n ` `/` `i ` ` ` `product ` `=` `product ` `*` `i ` ` ` `i ` `=` `i ` `+` `1` ` ` `if` `n >` `=` `2` `: ` ` ` `product ` `=` `product ` `*` `n ` ` ` `return` `product ` ` ` `# Driver code ` ` ` `# Get the binary string ` `n ` `=` `20` `print` `(minimum(n)) ` ` ` `# This code is contributed ` `# by Shashank_Sharma ` |

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## C#

`// C# implementation of the above approach ` `class` `GFG ` `{ ` ` ` `// function to return the product of ` `// distinct prime factors of a number ` `static` `int` `minimum(` `int` `n) ` `{ ` ` ` `int` `product = 1; ` ` ` ` ` `// find distinct prime ` ` ` `for` `(` `int` `i = 2; i * i <= n; i++) ` ` ` `{ ` ` ` `if` `(n % i == 0) ` ` ` `{ ` ` ` `while` `(n % i == 0) ` ` ` `n = n / i; ` ` ` `product = product * i; ` ` ` `} ` ` ` `} ` ` ` `if` `(n >= 2) ` ` ` `product = product * n; ` ` ` ` ` `return` `product; ` `} ` ` ` `// Driver code ` `static` `void` `Main() ` `{ ` ` ` `int` `n = 20; ` ` ` `System.Console.WriteLine(minimum(n)); ` `} ` `} ` ` ` `// This code is contributed by mits ` |

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## PHP

`<?php ` `// PHP implementation of the ` `// above approach ` ` ` `// function to return the product of ` `// distinct prime factors of a number ` `function` `minimum(` `$n` `) ` `{ ` ` ` `$product` `= 1; ` ` ` ` ` `// find distinct prime ` ` ` `for` `(` `$i` `= 2; ` `$i` `* ` `$i` `<= ` `$n` `; ` `$i` `++) ` ` ` `{ ` ` ` `if` `(` `$n` `% ` `$i` `== 0) ` ` ` `{ ` ` ` `while` `(` `$n` `% ` `$i` `== 0) ` ` ` `$n` `= ` `$n` `/ ` `$i` `; ` ` ` `$product` `= ` `$product` `* ` `$i` `; ` ` ` `} ` ` ` `} ` ` ` `if` `(` `$n` `>= 2) ` ` ` `$product` `= ` `$product` `* ` `$n` `; ` ` ` ` ` `return` `$product` `; ` `} ` ` ` `// Driver code ` `$n` `= 20; ` `echo` `minimum(` `$n` `),` `"\n"` `; ` ` ` `// This code is contributed by ANKITRAI1 ` `?> ` |

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**Output:**

10

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