Find the Smallest number that divides X^X

Difficulty Level : Basic
Last Updated : 23 Mar, 2021

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))

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes

Related Articles

C++

Java

Python3

C#

Javascript