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Even Perfect Number

  • Difficulty Level : Easy
  • Last Updated : 13 Sep, 2021

Given an even number N, the task is to check whether it is a Perfect number or not without finding its divisors.

In number theory, an Even Perfect Number is a positive integer which is even or that is equal to the sum of its positive divisors, excluding the number itself.

An even perfect number can be represented as P * (P + 1) / 2 where P is Mersenne Prime.

A Mersenne Prime is a prime number of form 2q – 1 where q is also a prime number.
For example: if N = 6, 
If we choose q to be 2 (prime number) then mersenne prime (P) is 22 – 1 = 3. 
Therefore, the Even perfect number formed by the formula is 3 * (3 + 1) / 2 = 6.

Examples:

Input: N = 6
Output: Yes
Explanation:
The integer 6 can be written as  6 = 1 + 2 + 3. Hence, its perfect number.



Input: N  =156
Output: No
Explanation:
The integer 156 cannot be written as a sum of its divisors. Hence, its not a perfect number.

Approach: 
 

  1. Find the square root of the given number to get a number close to 2q – 1.
  2. Find q-1 from the square root of the number and then check whether 2q-1 * (2q-1) gives the number entered. If not then it is not a perfect number, otherwise continue.
  3. Check whether q is prime or not. If it is not prime then 2q-1 cannot be prime and subsequently check whether 2q-1 is prime.
  4. If all the above conditions hold true then it is an even perfect number otherwise not.

Below is the implementation of the above approach:
 

C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
bool isPrime(long n);
 
// Function to check for perfect number
void check(long num)
{
     
    // Find a number close to 2^q-1
    long root = (long)sqrt(num);
 
    // Calculate q-1
    long poww = (long)(log(root) / log(2));
 
    // Condition of perfect number
    if (num == (long)(pow(2, poww) *
                    (pow(2, poww + 1) - 1)))
    {
 
        // Check whether q is prime or not
        if (isPrime(poww + 1))
        {
             
            // Check whether 2^q - 1 is a
            // prime number or not
            if (isPrime((long)pow(2,
                poww + 1) - 1))
                cout << "Yes" << endl;
            else
                cout << "No" << endl;
        }
        else
            cout << "No" << endl;
    }
    else
        cout << "No" << endl;
}
 
// Function to check for prime number
bool isPrime(long n)
{
    if (n <= 1)
        return false;
 
    // Check whether it is equal to 2 or 3
    else if (n == 2 || n == 3)
        return true;
 
    else
    {
         
        // Check if it can be divided by 2
        // and 3 then it is not prime number
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        // Check whether the given number be
        // divide by other prime numbers
        for(long i = 5; i <= sqrt(n); i += 6)
        {
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        }
        return true;
    }
}
 
// Driver Code
int main()
{
    long num = 6;
     
    check(num);
     
    return 0;
}
 
// This code is contributed by rutvik_56

Java




// Java program for the above approach
 
class GFG {
 
    // Function to check for perfect number
    private static void check(long num)
    {
        // Find a number close to 2^q-1
        long root = (long)Math.sqrt(num);
 
        // Calculate q-1
        long pow
            = (long)(Math.log(root)
                    / Math.log(2));
 
        // Condition of perfect number
        if (num
            == (long)(Math.pow(2, pow)
                    * (Math.pow(2, pow + 1) - 1))) {
 
            // Check whether q is prime or not
            if (isPrime(pow + 1)) {
 
                // Check whether 2^q - 1 is a
                // prime number or not
                if (isPrime(
                        (long)Math.pow(
                            2, pow + 1)
                        - 1))
                    System.out.println("Yes");
 
                else
                    System.out.println("No");
            }
            else
                System.out.println("No");
        }
        else
            System.out.println("No");
    }
 
    // Function to check for prime number
    public static boolean isPrime(long n)
    {
        if (n <= 1)
            return false;
 
        // Check whether it is equal to 2 or 3
        else if (n == 2 || n == 3)
            return true;
 
        else {
            // Check if it can be divided by 2
            // and 3 then it is not prime number
            if (n % 2 == 0 || n % 3 == 0)
                return false;
 
            // Check whether the given number be
            // divide by other prime numbers
            for (long i = 5;
                i <= Math.sqrt(n);
                i += 6) {
                if (n % i == 0
                    || n % (i + 2) == 0)
                    return false;
            }
            return true;
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        long num = 6;
        check(num);
    }
}

Python3




# Python3 program for the above approach
import math
 
# Function to check for perfect number
def check(num):
     
    # Find a number close to 2^q-1
    root = (int)(math.sqrt(num))
 
    # Calculate q-1
    poww = (int)(math.log(root) /
                 math.log(2))
 
    # Condition of perfect number
    if (num == (int)(pow(2, poww) *
                    (pow(2, poww + 1) - 1))):
 
        # Check whether q is prime or not
        if (isPrime(poww + 1)):
             
            # Check whether 2^q - 1 is a
            # prime number or not
            if (isPrime((int)(pow(2,
                poww + 1)) - 1)):
                print("Yes")
            else:
                print("No")
                 
        else:
            print("No")
    else:
        print("No")
 
# Function to check for prime number
def isPrime(n):
 
    if (n <= 1):
        return bool(False)
 
    # Check whether it is equal to 2 or 3
    elif (n == 2 or n == 3):
        return bool(True)
 
    else:
         
        # Check if it can be divided by 2
        # and 3 then it is not prime number
        if (n % 2 == 0 or n % 3 == 0):
            return bool(False)
 
        # Check whether the given number be
        # divide by other prime numbers
        for i in range(5, sqrt(n + 1) + 1, 6):
            if (n % i == 0 or n % (i + 2) == 0):
                return bool(False)
                 
        return bool(True)
 
# Driver Code        
num = 6
     
check(num)
 
# This code is contributed by divyeshrabadiya07

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check for perfect number
private static void check(long num)
{
     
    // Find a number close to 2^q-1
    long root = (long)Math.Sqrt(num);
 
    // Calculate q-1
    long pow = (long)(Math.Log(root) /
                    Math.Log(2));
 
    // Condition of perfect number
    if (num == (long)(Math.Pow(2, pow) *
                    (Math.Pow(2, pow + 1) - 1)))
    {
         
        // Check whether q is prime or not
        if (isPrime(pow + 1))
        {
 
            // Check whether 2^q - 1 is a
            // prime number or not
            if (isPrime((long)Math.Pow(2, pow + 1) - 1))
                Console.WriteLine("Yes");
 
            else
                Console.WriteLine("No");
        }
        else
            Console.WriteLine("No");
    }
    else
        Console.WriteLine("No");
}
 
// Function to check for prime number
public static bool isPrime(long n)
{
    if (n <= 1)
        return false;
 
    // Check whether it is equal to 2 or 3
    else if (n == 2 || n == 3)
        return true;
 
    else
    {
         
        // Check if it can be divided by 2
        // and 3 then it is not prime number
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        // Check whether the given number be
        // divide by other prime numbers
        for(long i = 5;
                i <= Math.Sqrt(n);
                i += 6)
        {
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        }
        return true;
    }
}
 
// Driver code
public static void Main(String []args)
{
    long num = 6;
    check(num);
}
}
 
// This code is contributed by amal kumar choubey

Javascript




<script>
// JavaScript program for the above approach
 
    // Function to check for perfect number
    function check(num)
    {
        // Find a number close to 2^q-1
        let root = Math.floor(Math.sqrt(num));
   
        // Calculate q-1
        let pow
            = Math.floor(Math.log(root)
                    / Math.log(2));
   
        // Condition of perfect number
        if (num
            == Math.floor(Math.pow(2, pow)
                    * (Math.pow(2, pow + 1) - 1))) {
   
            // Check whether q is prime or not
            if (isPrime(pow + 1)) {
   
                // Check whether 2^q - 1 is a
                // prime number or not
                if (isPrime(
                        Math.floor(Math.pow(
                            2, pow + 1) )
                        - 1))
                    document.write("Yes");
   
                else
                    document.write("No");
            }
            else
                document.write("No");
        }
        else
            document.write("No");
    }
   
    // Function to check for prime number
    function isPrime(n)
    {
        if (n <= 1)
            return false;
   
        // Check whether it is equal to 2 or 3
        else if (n == 2 || n == 3)
            return true;
   
        else {
            // Check if it can be divided by 2
            // and 3 then it is not prime number
            if (n % 2 == 0 || n % 3 == 0)
                return false;
   
            // Check whether the given number be
            // divide by other prime numbers
            for (let i = 5;
                i <= Math.floor(Math.sqrt(n));
                i += 6) {
                if (n % i == 0
                    || n % (i + 2) == 0)
                    return false;
            }
            return true;
        }
    }
 
// Driver Code   
     
    let num = 6;
    check(num);
                   
</script>
Output: 
Yes

 

Time Complexity: O(N1/4)
Auxiliary Space: O(1)




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