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.

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 5

Time Complexity: O(sqrt(X))

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