Skip to content
Related Articles

Related Articles

Improve Article

Multiply perfect number

  • Last Updated : 16 Jul, 2021

A number N is said to be Multiply-perfect numbers if N divides sigma(N), where sigma(N) = sum of all divisors of N.
The first few Multiply-perfect numbers are: 
 

1, 6, 28, 120, 496, 672, …….. 
 

 

Check if N is a Multiply-perfect number

Given a number N, the task is to find if this number is Multiply-perfect number or not. 
Examples: 
 

Input: N = 120 
Output: YES 
Explanation: 
Sum of 120’s divisors is 1 + 2 + 3 + 4 + 5 + 6 + 8 + 10 + 12 + 15 + 20 + 24 + 30 + 40 + 60 + 120 = 360 and 120 divides 360. 
Therefore, 120 is a Multiply-perfect number.
Input: N = 32 
Output: No 
 



 

Approach: For a number N to be Multiply-perfect number, the following condition should hold true: sigma(N) % N = 0, where sigma(N) = sum of all divisors of N. Therefore, we will find sum of all divisors of N and check if it is divisible by N or not. If divisible print “Yes” else print “No.
Below is the implementation of the above approach:
 

C++




// C++ implementation of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// sum of divisors
int getSum(int n)
{
    int sum = 0;
 
    // Note that this loop
    // runs till square root of N
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
 
            // If divisors are equal,
            // take only one of them
            if (n / i == i)
                sum = sum + i;
 
            // Otherwise take both
            else {
                sum = sum + i;
                sum = sum + (n / i);
            }
        }
    }
 
    return sum;
}
 
// Function to check Multiply-perfect number
bool MultiplyPerfectNumber(int n)
{
    if (getSum(n) % n == 0)
        return true;
    else
        return false;
}
 
// Driver code
int main()
{
 
    int n = 28;
    if (MultiplyPerfectNumber(n)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java implementation of the above approach
class GFG{
     
// Function to find the
// sum of divisors
static int getSum(int n)
{
    int sum = 0;
 
    // Note that this loop
    // runs till square root of N
    for(int i = 1; i <= Math.sqrt(n); i++)
    {
       if (n % i == 0)
       {
            
           // If divisors are equal,
           // take only one of them
           if (n / i == i)
               sum = sum + i;
            
           // Otherwise take both
           else
           {
               sum = sum + i;
               sum = sum + (n / i);
           }
       }
    }
    return sum;
}
 
// Function to check Multiply-perfect number
static boolean MultiplyPerfectNumber(int n)
{
    if (getSum(n) % n == 0)
        return true;
    else
        return false;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 28;
     
    if (MultiplyPerfectNumber(n))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Ritik Bansal

Python3




# Python3 implementation of the above approach
import math
 
# Function to find the
# sum of divisors
def getSum(n):
 
    sum1 = 0;
 
    # Note that this loop
    # runs till square root of N
    for i in range(1, int(math.sqrt(n))):
        if (n % i == 0):
 
            # If divisors are equal,
            # take only one of them
            if (n // i == i):
                sum1 = sum1 + i;
 
            # Otherwise take both
            else:
                sum1 = sum1 + i;
                sum1 = sum1 + (n // i);
             
    return sum1;
 
# Function to check Multiply-perfect number
def MultiplyPerfectNumber(n):
 
    if (getSum(n) % n == 0):
        return True;
    else:
        return False;
 
# Driver code
n = 28;
if (MultiplyPerfectNumber(n)):
    print("Yes");
else:
    print("No");
 
# This code is contributed by Code_Mech

C#




// C# implementation of the above approach
using System;
class GFG{
     
// Function to find the
// sum of divisors
static int getSum(int n)
{
    int sum = 0;
 
    // Note that this loop
    // runs till square root of N
    for(int i = 1; i <= Math.Sqrt(n); i++)
    {
       if (n % i == 0)
       {
            
           // If divisors are equal,
           // take only one of them
           if (n / i == i)
               sum = sum + i;
            
           // Otherwise take both
           else
           {
               sum = sum + i;
               sum = sum + (n / i);
           }
       }
    }
    return sum;
}
 
// Function to check Multiply-perfect number
static bool MultiplyPerfectNumber(int n)
{
    if (getSum(n) % n == 0)
        return true;
    else
        return false;
}
 
// Driver code
public static void Main()
{
    int n = 28;
     
    if (MultiplyPerfectNumber(n))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Code_Mech

Javascript




<script>
// Javascript implementation of the above approach
 
    // Function to find the
    // sum of divisors
    function getSum( n)
    {
        let sum = 0;
 
        // Note that this loop
        // runs till square root of N
        for ( i = 1; i <= Math.sqrt(n); i++)
        {
            if (n % i == 0)
            {
 
                // If divisors are equal,
                // take only one of them
                if (n / i == i)
                    sum = sum + i;
 
                // Otherwise take both
                else {
                    sum = sum + i;
                    sum = sum + (n / i);
                }
            }
        }
        return sum;
    }
 
    // Function to check Multiply-perfect number
    function MultiplyPerfectNumber( n) {
        if (getSum(n) % n == 0)
            return true;
        else
            return false;
    }
 
    // Driver code
      
    let n = 28;
    if (MultiplyPerfectNumber(n)) {
        document.write("Yes");
    } else {
        document.write("No");
    }
 
// This code is contributed by Rajput-Ji
</script>
Output: 
Yes

 

Time Complexity: O(N1/2)

Auxiliary Space: O(1)

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

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 experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :