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Powerful Number
  • Difficulty Level : Basic
  • Last Updated : 14 May, 2018

A number n is said to be Powerful Number if for every prime factor p of it, p2 also divides it. For example:- 36 is a powerful number. It is divisible by both 3 and square of 3 i.e, 9.

The first few Powerful Numbers are:
1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64 ….

Given a number n, our task is to check if this is powerful or not.
Examples :

Input: 27
Output: YES

Input: 32
Output: YES

Input: 12
Output: NO

The idea is based on the fact that if a number n is powerful, then all prime factors of it and their squares should be divisible by n. We find all prime factors of given number. And for every prime factor, we find the highest power of it that divides n. If we find a prime factor whose highest dividing power is 1, we return false. If highest dividing power of all prime factors is more than 1, we return true.



Below is implementation of above idea.

C++




// C++ program to find if a number is powerful or not.
#include <bits/stdc++.h>
using namespace std;
  
// function to check if the number is powerful
bool isPowerful(int n)
{
    // First divide the number repeatedly by 2
    while (n % 2 == 0) {
        int power = 0;
        while (n % 2 == 0) {
            n /= 2;
            power++;
        }
  
        // If only 2^1 divides n (not higher powers),
        // then return false
        if (power == 1)
            return false;
    }
  
    // if n is not a power of 2 then this loop will execute
    // repeat above process
    for (int factor = 3; factor <= sqrt(n); factor += 2) {
        // Find highest power of "factor" that divides n
        int power = 0;
        while (n % factor == 0) {
            n = n / factor;
            power++;
        }
  
        // If only factor^1 divides n (not higher powers),
        // then return false
        if (power == 1)
            return false;
    }
  
    // n must be 1 now if it is not a prime numenr.
    // Since prime numbers are not powerful, we return
    // false if n is not 1.
    return (n == 1);
}
  
// Driver program to test above function
int main()
{
    isPowerful(20) ? cout << "YES\n" : cout << "NO\n";
    isPowerful(27) ? cout << "YES\n" : cout << "NO\n";
  
    return 0;
}


Java




// Java program to find if a
// number is powerful or not.
  
class GFG {
    // function to check if the
    // number is powerful
    static boolean isPowerful(int n)
    {
        // First divide the number
        // repeatedly by 2
        while (n % 2 == 0) {
            int power = 0;
            while (n % 2 == 0) {
                n /= 2;
                power++;
            }
  
            // If only 2^1 divides n (not higher powers),
            // then return false
            if (power == 1)
                return false;
        }
  
        // if n is not a power of 2 then this loop will execute
        // repeat above process
        for (int factor = 3; factor <= Math.sqrt(n); factor += 2) {
            // Find highest power of "factor" that divides n
            int power = 0;
            while (n % factor == 0) {
                n = n / factor;
                power++;
            }
  
            // If only factor^1 divides n (not higher powers),
            // then return false
            if (power == 1)
                return false;
        }
  
        // n must be 1 now if it is not a prime numenr.
        // Since prime numbers are not powerful, we return
        // false if n is not 1.
        return (n == 1);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        if (isPowerful(20))
            System.out.print("YES\n");
        else
            System.out.print("NO\n");
        if (isPowerful(27))
            System.out.print("YES\n");
        else
            System.out.print("NO\n");
    }
}
  
// This code is contributed by Anant Agarwal.


Python3




# Python program to find
# if a number is powerful or not.
import math
  
# function to check if
# the number is powerful
def isPowerful(n):
  
    # First divide the number repeatedly by 2
    while (n % 2 == 0):
  
        power = 0
        while (n % 2 == 0):
          
            n = n//2 
            power = power + 1
          
           
        # If only 2 ^ 1 divides
        # n (not higher powers),
        # then return false
        if (power == 1):
            return False
      
   
    # if n is not a power of 2
    # then this loop will execute
    # repeat above process
    for factor in range(3, int(math.sqrt(n))+1, 2):
      
        # Find highest power of
        # "factor" that divides n
        power = 0
        while (n % factor == 0):
          
            n = n//factor
            power = power + 1
          
   
        # If only factor ^ 1 divides
        # n (not higher powers),
        # then return false
        if (power == 1):
            return false
      
   
     # n must be 1 now if it
     # is not a prime numenr.
     # Since prime numbers are
     # not powerful, we return
     # false if n is not 1.
    return (n == 1)
  
# Driver code
  
print("YES" if isPowerful(20) else "NO")
print("YES" if isPowerful(27) else "NO")
  
  
# This code is contributed
# by Anant Agarwal.


C#




// C# program to find if a
// number is powerful or not.
using System;
  
class GFG {
  
    // function to check if the
    // number is powerful
    static bool isPowerful(int n)
    {
        // First divide the number
        // repeatedly by 2
        while (n % 2 == 0) {
            int power = 0;
            while (n % 2 == 0) {
                n /= 2;
                power++;
            }
  
            // If only 2^1 divides n
            // (not higher powers),
            // then return false
            if (power == 1)
                return false;
        }
  
        // if n is not a power of 2 then
        // this loop will execute repeat
        // above process
        for (int factor = 3; factor <= Math.Sqrt(n); factor += 2) {
            // Find highest power of "factor"
            // that divides n
            int power = 0;
            while (n % factor == 0) {
                n = n / factor;
                power++;
            }
  
            // If only factor^1 divides n
            // (not higher powers), then
            // return false
            if (power == 1)
                return false;
        }
  
        // n must be 1 now if it is not
        // a prime numenr.
        // Since prime numbers are not
        // powerful, we return false if
        // n is not 1.
        return (n == 1);
    }
  
    // Driver code
    public static void Main()
    {
        if (isPowerful(20))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
  
        if (isPowerful(27))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}
  
// This code is contributed by vt_m.


PHP




<?php
// PHP program to find if a
// number is powerful or not
  
// function to check if 
// the number is powerful
function isPowerful($n)
{
    // First divide the 
    // number repeatedly by 2
    while ($n % 2 == 0)
    {
        $power = 0;
        while ($n % 2 == 0)
        {
            $n /= 2;
            $power++;
        }
          
        // If only 2^1 divides 
        // n (not higher powers),
        // then return false
        if ($power == 1)
        return false;
    }
  
    // if n is not a power of 2 
    // then this loop will execute
    // repeat above process
    for ($factor = 3; 
         $factor <= sqrt($n); 
         $factor += 2)
    {
        // Find highest power of 
        // "factor" that divides n
        $power = 0;
        while ($n % $factor == 0)
        {
            $n = $n / $factor;
            $power++;
        }
  
        // If only factor^1 divides 
        // n (not higher powers),
        // then return false
        if ($power == 1)
        return false;
    }
  
    // n must be 1 now if it is
    // not a prime number. Since 
    // prime numbers are not powerful, 
    // we return false if n is not 1.
    return ($n == 1);
}
  
// Driver Code
$d = isPowerful(20) ? 
            "YES\n"
              "NO\n";
    echo $d;
$d = isPowerful(27) ? 
            "YES\n"
              "NO\n";
    echo $d;
  
// This code is contributed by ajit.     
?>



Output :

NO
YES

References:
https://en.wikipedia.org/wiki/Powerful_number

This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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