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:
- Find the square root of the given number to get a number close to 2q – 1.
- 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.
- 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.
- 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++
#include<bits/stdc++.h>
using namespace std;
bool isPrime( long n);
void check( long num)
{
long root = ( long ) sqrt (num);
long poww = ( long )( log (root) / log (2));
if (num == ( long )( pow (2, poww) *
( pow (2, poww + 1) - 1)))
{
if (isPrime(poww + 1))
{
if (isPrime(( long ) pow (2,
poww + 1) - 1))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
else
cout << "No" << endl;
}
else
cout << "No" << endl;
}
bool isPrime( long n)
{
if (n <= 1)
return false ;
else if (n == 2 || n == 3)
return true ;
else
{
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( long i = 5; i <= sqrt (n); i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
return false ;
}
return true ;
}
}
int main()
{
long num = 6;
check(num);
return 0;
}
|
Java
class GFG {
private static void check( long num)
{
long root = ( long )Math.sqrt(num);
long pow
= ( long )(Math.log(root)
/ Math.log( 2 ));
if (num
== ( long )(Math.pow( 2 , pow)
* (Math.pow( 2 , pow + 1 ) - 1 ))) {
if (isPrime(pow + 1 )) {
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" );
}
public static boolean isPrime( long n)
{
if (n <= 1 )
return false ;
else if (n == 2 || n == 3 )
return true ;
else {
if (n % 2 == 0 || n % 3 == 0 )
return false ;
for ( long i = 5 ;
i <= Math.sqrt(n);
i += 6 ) {
if (n % i == 0
|| n % (i + 2 ) == 0 )
return false ;
}
return true ;
}
}
public static void main(String args[])
{
long num = 6 ;
check(num);
}
}
|
Python3
import math
def check(num):
root = ( int )(math.sqrt(num))
poww = ( int )(math.log(root) /
math.log( 2 ))
if (num = = ( int )( pow ( 2 , poww) *
( pow ( 2 , poww + 1 ) - 1 ))):
if (isPrime(poww + 1 )):
if (isPrime(( int )( pow ( 2 ,
poww + 1 )) - 1 )):
print ( "Yes" )
else :
print ( "No" )
else :
print ( "No" )
else :
print ( "No" )
def isPrime(n):
if (n < = 1 ):
return bool ( False )
elif (n = = 2 or n = = 3 ):
return bool ( True )
else :
if (n % 2 = = 0 or n % 3 = = 0 ):
return bool ( False )
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 )
num = 6
check(num)
|
C#
using System;
using System.Collections.Generic;
class GFG{
private static void check( long num)
{
long root = ( long )Math.Sqrt(num);
long pow = ( long )(Math.Log(root) /
Math.Log(2));
if (num == ( long )(Math.Pow(2, pow) *
(Math.Pow(2, pow + 1) - 1)))
{
if (isPrime(pow + 1))
{
if (isPrime(( long )Math.Pow(2, pow + 1) - 1))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
else
Console.WriteLine( "No" );
}
else
Console.WriteLine( "No" );
}
public static bool isPrime( long n)
{
if (n <= 1)
return false ;
else if (n == 2 || n == 3)
return true ;
else
{
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( long i = 5;
i <= Math.Sqrt(n);
i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
return false ;
}
return true ;
}
}
public static void Main(String []args)
{
long num = 6;
check(num);
}
}
|
Javascript
<script>
function check(num)
{
let root = Math.floor(Math.sqrt(num));
let pow
= Math.floor(Math.log(root)
/ Math.log(2));
if (num
== Math.floor(Math.pow(2, pow)
* (Math.pow(2, pow + 1) - 1))) {
if (isPrime(pow + 1)) {
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 isPrime(n)
{
if (n <= 1)
return false ;
else if (n == 2 || n == 3)
return true ;
else {
if (n % 2 == 0 || n % 3 == 0)
return false ;
for (let i = 5;
i <= Math.floor(Math.sqrt(n));
i += 6) {
if (n % i == 0
|| n % (i + 2) == 0)
return false ;
}
return true ;
}
}
let num = 6;
check(num);
</script>
|
Time Complexity: O(N1/4)
Auxiliary Space: O(1)
Last Updated :
13 Sep, 2021
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