We can set a loop to iterate from 1 to n-1.

In every iteration we check if n is divisible by i, if it is we add it to our sum.

After the loop is over, we check whether the given number is equal to our sum, if it is then we the given number is a perfect number, otherwise not.

Code

#include <bits/stdc++.h>
using namespace std;

// Function to check if a number is a perfect number
bool isPerfect(int n) {
    // Variable to store the sum of divisors
    int sum = 0;

    // Loop to find all divisors of the number and calculate their sum
    for (int i = 1; i <= n - 1; i++) {
        // If i is a divisor of n
        if (n % i == 0)
            sum = sum + i;  // Add i to the sum of divisors
    }

    // If the sum of divisors is equal to n, it is a perfect number
    if (sum == n)
        return true;
    else
        return false;  // Otherwise, it's not a perfect number
}

int main() {
    // Check for different numbers if they are perfect numbers
    bool ex1 = isPerfect(6);   // Check if 6 is a perfect number
    bool ex2 = isPerfect(15);  // Check if 15 is a perfect number
    bool ex3 = isPerfect(28);  // Check if 28 is a perfect number

    // Output the result for each example
    if (ex1 == true) {
        cout << "6 is 