Find if the given number is a Perfect Number

        public static bool IsPerfectNumber(int n)
        {
            int sum = 0;
            Console.Write("Factors:");
            for (int i = 1; i < n; i++)
            {
                if (n % i == 0)
                {
                    Console.Write($"{i} ");
                    sum = sum + i;
                }
            }

            Console.WriteLine($"Sum: {sum}");

            return sum == n ? true : false;
        }
        

How it works:

While calculating the factors, we exclude the input number from the loop (the reason why the condition is i < n and not i <=n) and for each time the counter is a factor of the input number we add it to the sum. Finally, we compare the sum with the input number for equality and the result tells us if its a Perfect Number or not.

Number: 6
Factors:1 2 3 Sum: 6
True