Perfect Number - Solution & Explanation

The main idea is that divisors come in pairs, and for each pair (d, n/d), if d is less than the square root of n, then d and n/d are both divisors of n. This approach allows us to only iterate up to the square root of n, making it more efficient.

The function first checks if num is less than or equal to 1, returning false if so, as there's no perfect number <= 1. Then it initializes sum with 1 since 1 is a divisor of every number except itself. We iterate through possible divisors up to the square root of num. If i is a divisor, we add it and its pair num/i to sum, avoiding double counting. Finally, we check if sum equals num.

This approach calculates the sum of all divisors excluding the number itself by checking each integer from 1 to n-1 to determine if it's a divisor, then sums these and checks if the sum equals n. This method is simple but inefficient for large numbers.

The C solution performs a simple loop through possible divisors from 1 to num - 1, summing any divisors found and checking if this total matches num.