Perfect Number

A number is a perfect number if is equal to sum of its proper divisors, that is, sum of its positive divisors excluding the number itself. Write a function to check if a given number is perfect or not. 
Examples:

Input: n = 15
Output: false
Divisors of 15 are 1, 3 and 5. Sum of 
divisors is 9 which is not equal to 15.
Input: n = 6
Output: true
Divisors of 6 are 1, 2 and 3. Sum of 
divisors is 6.

A Simple Solution is to go through every number from 1 to n-1 and check if it is a divisor. Maintain sum of all divisors. If sum becomes equal to n, then return true, else return false.
An Efficient Solution is to go through numbers till square root of n. If a number ‘i’ divides n, then add both ‘i’ and n/i to sum. 
Below is the implementation of efficient solution. 

C++

// C++ program to check if a given number is perfect
// or not
#include<iostream>
using namespace std;

// Returns true if n is perfect
bool isPerfect(long long int n)
{
    // To store sum of divisors
    long long int sum = 1;

    // Find all divisors and add them
    for (long long int i=2; i*i<=n; i++)
    {
        if (n%i==0)
        {
            if(i*i!=n)
                sum = sum + i + n/i;
            else
                sum=sum+i;
        }
    }
     // If sum of divisors is equal to
     // n, then n is a perfect number
     if (sum == n && n != 1)
          return true;

     return false;
}

// Driver program
int main()
{
    cout << "Below are all perfect numbers till 10000\n";
    for (int n =2; n<10000; n++)
        if (isPerfect(n))
            cout << n << " is a perfect number\n";

    return 0;
}

Java

// Java program to check if a given
// number is perfect or not
import java.io.*;
public class GFG
{

// Returns true if n is perfect
static boolean isPerfect(int n)
{
    // To store sum of divisors
    int sum = 1;

    // Find all divisors and add them
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i==0)
        {
            if(i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
    // If sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;

    return false;
}

// Driver program
public static void main (String[] args)
{
    System.out.println("Below are all perfect" +
                                "numbers till 10000");
    for (int n = 2; n < 10000; n++)
        if (isPerfect(n))
            System.out.println( n +
                    " is a perfect number");
}
}

// This code is contributed by mits

Python3

# Python3 code to check if a given
# number is perfect or not

# Returns true if n is perfect
def isPerfect( n ):

    # To store sum of divisors
    sum = 1

    # Find all divisors and add them
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1

    # If sum of divisors is equal to
    # n, then n is a perfect number

    return (True if sum == n and n!=1 else False)

# Driver program
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to check if a given
// number is perfect or not

class GFG
{

// Returns true if n is perfect
static bool isPerfect(int n)
{
    // To store sum of divisors
    int sum = 1;

    // Find all divisors and add them
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i==0)
        {
            if(i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
    // If sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;

    return false;
}

// Driver program
static void Main()
{
    System.Console.WriteLine("Below are all perfect" +
                                "numbers till 10000");
    for (int n = 2; n < 10000; n++)
        if (isPerfect(n))
            System.Console.WriteLine( n +
                    " is a perfect number");
}
}

// This code is contributed by chandan_jnu

Javascript

<script>

// Javascript program to check if a given number is perfect
// or not

// Returns true if n is perfect
function isPerfect(n)
{
    // To store sum of divisors
    sum = 1;

    // Find all divisors and add them
    for (let i=2; i*i<=n; i++)
    {
        if (n%i==0)
        {
            if(i*i!=n)
                sum = sum + i + n/i;
            else
                sum=sum+i;
        }
    }
    // If sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;

    return false;
}

// Driver program

    document.write("Below are all perfect numbers till 10000" + "<br>");
    for (let n =2; n<10000; n++)
        if (isPerfect(n))
            document.write(n + " is a perfect number" + "<br>");

// This code is contributed by Mayank Tyagi

</script>

PHP

<?php
// PHP program to check if a given number
// is perfect or not

// Returns true if n is perfect
function isPerfect($n)
{
    // To store sum of divisors
    $sum = 1;

    // Find all divisors and add them
    for ($i = 2; $i * $i <= $n; $i++)
    {
        if ($n % $i == 0)
        {
            if($i * $i != $n)
                $sum = $sum + $i + (int)($n / $i);
            else
                $sum = $sum + $i;
        }
    }
    // If sum of divisors is equal to
    // n, then n is a perfect number
    if ($sum == $n && $n != 1)
        return true;

    return false;
}

// Driver Code
echo "Below are all perfect numbers till 10000\n";
for ($n = 2; $n < 10000; $n++)
    if (isPerfect($n))
        echo "$n is a perfect number\n";

// This code is contributed by mits
?>

Output

Below are all perfect numbers till 10000
6 is a perfect number
28 is a perfect number
496 is a perfect number
8128 is a perfect number

Time Complexity: O(?n)
Auxiliary Space: O(1), since no extra space has been taken.
Below are some interesting facts about Perfect Numbers: 
1) Every even perfect number is of the form 2^p?1(2^p ? 1) where 2^p ? 1 is prime. 
2) It is unknown whether there are any odd perfect numbers.