Find the Smallest number that divides X^X

Given a number X, find the Smallest Number, which is greater than 1, that divides X^X. 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. 

Approach:
The main observation of this problem is that if a number P divides X, then it also divides X^X, so we don’t need to calculate the value of X^X. 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 <bits/stdc++.h>
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

Javascript

<script>

    // Javascript implementation of above approach

    // Function to find the required smallest number
    function SmallestDiv(n)
    {

        for (let 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;
    }

    let X = 385;

    let ans = SmallestDiv(X);
    document.write(ans + "</br>");

</script>

Output: 5

Time Complexity: O(sqrt(X))