Perfect Number (original) (raw)

Last Updated : 23 Jul, 2025

A number is a perfect number if it is equal to the sum of its proper divisors, that is, the sum of its positive divisors excluding the number itself. Find whether a given positive integer n is perfect or not.
**Examples:

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

**Input: n = 6
**Output: true
**Explanation: Divisors of 6 are 1, 2 and 3. Sum of divisors is 6.

Try It Yourself

Table of Content

• [Naive Approach] Divisor Sum Method - O(n) Time and O(1) Space
• [Expected Approach] Optimized Divisor Search Method - O(sqrt n) Time and O(1) Space

[Naive Approach] Divisor Sum Method - O(n) Timeand O(1)Space

A simple Solutionis 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.

C++ `

#include using namespace std;

// Function to check if the number is perfect bool isPerfect(int n) { int sum = 0; for (int i = 1; i < n; i++) { if (n % i == 0) { sum += i; } } return sum == n; }

int main() { int n = 15; cout << (isPerfect(n) ?"true" : "false"); return 0; }

C

#include <stdio.h> #include <stdbool.h>

// Function to check if the number is perfect bool isPerfect(int n) { int sum = 0; for (int i = 1; i < n; i++) { if (n % i == 0) { sum += i; } } return sum == n; }

int main() { int n = 15; printf(isPerfect(n) ? "true" : "false"); return 0; }

Java

public class GFG {

// Function to check if the number is perfect
public static boolean isPerfect(int n) {
    int sum = 0;
    for (int i = 1; i < n; i++) {
        if (n % i == 0) {
            sum += i;
        }
    }
    return sum == n;
}

public static void main(String[] args) {
    int n = 15;
    System.out.println(isPerfect(n) ? "true" : "false");
}

}

Python

Function to check if the number is perfect

def isPerfect(n): sum = 0 for i in range(1, n): if n % i == 0: sum += i return sum == n

if name == 'main': n = 15 print("true" if isPerfect(n) else "false")

C#

using System;

class GFG { // Function to check if the number is perfect static bool isPerfect(int n) { int sum = 0; for (int i = 1; i < n; i++) { if (n % i == 0) { sum += i; } } return sum == n; }

static void Main() {
    int n = 15;
    Console.WriteLine(isPerfect(n) ? "true" : "false");
}

}

JavaScript

// Function to check if the number is perfect function isPerfect(n) { let sum = 0; for (let i = 1; i < n; i++) { if (n % i === 0) { sum += i; } } return sum === n; }

// Driver Code let n = 15; console.log(isPerfect(n) ? "true" : "false");

`

[Expected Approach] Optimized Divisor Search Method - O(sqrt n) Timeand O(1)Space

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' .

C++ `

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

// Returns true if n is perfect 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;

}

int main() { int n = 15; cout <<(isPerfect(n) ?"true" : "false");

return 0;

}

C

#include <stdio.h> #include <stdbool.h>

// Returns true if n is perfect 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;

}

int main() { int n = 15; printf("%s", isPerfect(n) ? "true" : "false");

return 0;

}

Java

public class GFG {

// Returns true if n is perfect
public 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
    return sum == n && n != 1;
}

public static void main(String[] args) {
    int n = 15;
    System.out.println(isPerfect(n) ? "true" : "false");
}

}

Python

def isPerfect(n):

# To store sum of divisors
sum = 1

# Find all divisors and add them
for i in range(2, int(n**0.5) + 1):
    if n % i == 0:
        if i * i != n:
            sum += i + n // i
        else:
            sum += i
            
# If sum of divisors is equal to
# n, then n is a perfect number
return sum == n and n != 1

if name == 'main': n = 15 print("true" if isPerfect(n) else "false")

C#

using System; class GfG {

// Returns true if n is perfect
public 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 += i + n / i;
            else
                sum += i;
        }
    }
    
    // If sum of divisors is equal to
    // n, then n is a perfect number
    return sum == n && n != 1;
}

public static void Main() {
    int n = 15;
    Console.WriteLine(IsPerfect(n) ? "true" : "false");
}

}

JavaScript

// Returns true if n is perfect function isPerfect(n) {

// To store sum of divisors
let 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 += i + n / i;
        else
            sum += i;
    }
}

// If sum of divisors is equal to
// n, then n is a perfect number
return sum === n && n !== 1;

}

// Driver program let n = 15; console.log(isPerfect(n) ? "true" : "false");

`

**Interesting facts About Perfect Number

1. Every even perfect number is of the form 2p−1⋅(2p−1) , where 2p−1 is a prime number.
2. It is unknown whether there are any odd perfect numbers.