# Find the Smallest number that divides X^X

Given a number X, find the Smallest Number, which is greater than 1, that divides XX. In the given question, X is always presumed to be greater than 1.

Examples:

Input: X = 6
Output: 2
Explanation: As, 66 is equal to 46656, which is divisible by 2 and it’s the smallest among all its divisors.

Input: X = 3
Output: 3
Explanation: As, 33 is equal to 27, which is divisible by 3 and it’s the smallest among all its divisors.

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

Approach:
The main observation of this problem is that if a number P divides X, then it also divides XX, so we don’t need to calculate the value of XX. What we need to do is to find the smallest number that divides X which will always be a Prime Number.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the required smallest number ` `int` `SmallestDiv(``int` `n) ` `{ ` `  `  `    ``for` `(``int` `i = 2; i * i <= n; i++) { ` `        ``// Finding smallest number that divides n ` `        ``if` `(n % i == 0) { ` ` `  `            ``// i divides n and return this  ` `            ``// value immediately ` `            ``return` `i; ` `        ``} ` `    ``} ` `  `  `    ``// If n is a prime number then answer should be n, ` `    ``// As we can't take 1 as our answer. ` `    ``return` `n; ` `} ` `  `  `// Driver Code ` `int` `main() ` `{ ` `  `  `    ``int` `X = 385; ` `  `  `    ``int` `ans = SmallestDiv(X); ` `    ``cout << ans << ``"\n"``; ` `  `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `class` `GFG{ ` ` `  `// Function to find the  ` `// required smallest number ` `static` `int` `SmallestDiv(``int` `n) ` `{ ` `    ``for``(``int` `i = ``2``; i * i <= n; i++) ` `    ``{ ` `         `  `       ``// Finding smallest number ` `       ``// that divides n ` `       ``if` `(n % i == ``0``) ` `       ``{ ` ` `  `           ``// i divides n and return this  ` `           ``// value immediately ` `           ``return` `i; ` `       ``} ` `    ``} ` ` `  `    ``// If n is a prime number then  ` `    ``// answer should be n, as we  ` `    ``// can't take 1 as our answer. ` `    ``return` `n; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `X = ``385``; ` `    ``int` `ans = SmallestDiv(X); ` `     `  `    ``System.out.print(ans + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by gauravrajput1 `

## Python3

 `# Python3 implementation of above approach ` ` `  `# Function to find the required smallest number ` `def` `SmallestDiv(n): ` `   `  `    ``i ``=` `2` `    ``while` `i ``*` `i <``=` `n: ` ` `  `        ``# Finding smallest number that divides n ` `        ``if` `(n ``%` `i ``=``=` `0``): ` `  `  `            ``# i divides n and return this  ` `            ``# value immediately ` `            ``return` `i ` `        ``i ``+``=` `1` `     `  `    ``# If n is a prime number then answer should be n, ` `    ``# As we can't take 1 as our answer. ` `    ``return` `n ` `   `  `# Driver Code ` `if` `__name__``=``=``"__main__"``: ` ` `  `    ``X ``=` `385` `    ``ans ``=` `SmallestDiv(X) ` ` `  `    ``print``(ans) ` ` `  `# This code is contributed by Yash_R `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG { ` `     `  `// Function to find the  ` `// required smallest number ` `static` `int` `SmallestDiv(``int` `n) ` `{ ` `    ``for``(``int` `i = 2; i * i <= n; i++) ` `    ``{ ` `         `  `       ``// Finding smallest number ` `       ``// that divides n ` `       ``if` `(n % i == 0) ` `       ``{ ` `            `  `           ``// i divides n and return this  ` `           ``// value immediately ` `           ``return` `i; ` `       ``} ` `    ``} ` ` `  `    ``// If n is a prime number then  ` `    ``// answer should be n, as we  ` `    ``// can't take 1 as our answer. ` `    ``return` `n; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `X = 385; ` `    ``int` `ans = SmallestDiv(X); ` `     `  `    ``Console.Write(ans + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by shivanisinghss2110 `

`Output: 5`

Time Complexity: O(sqrt(X))

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