C# Program to Check Perfect Number with Output and Explanation

A C# program to check a perfect number sums all its divisors except itself and compares the sum to the number using code like if (sum == number) Console.WriteLine("Perfect number"); else Console.WriteLine("Not perfect number");.

Examples

Input6

Output6 is a perfect number

Input28

Output28 is a perfect number

Input10

Output10 is not a perfect number

How to Think About It

To check if a number is perfect, find all numbers less than it that divide it evenly using the % operator. Add these divisors together. If the total equals the original number, it is perfect; otherwise, it is not.

Algorithm

1

Get the input number from the user

2

Initialize a sum variable to zero

3

Loop from 1 to number-1 and check if the current number divides the input evenly

4

If yes, add it to the sum

5

After the loop, compare sum with the input number

6

Print that the number is perfect if they are equal, else print not perfect

Code

csharp

using System;
class Program {
    static void Main() {
        Console.Write("Enter a number: ");
        int number = int.Parse(Console.ReadLine());
        int sum = 0;
        for (int i = 1; i < number; i++) {
            if (number % i == 0) sum += i;
        }
        if (sum == number) Console.WriteLine($"{number} is a perfect number");
        else Console.WriteLine($"{number} is not a perfect number");
    }
}

Output

Enter a number: 6
6 is a perfect number

Dry Run

Let's trace the input 6 through the code

1

Input number

User enters 6

2

Initialize sum

sum = 0

3

Loop and check divisors

Check i=1 to 5: divisors are 1, 2, 3 because 6 % i == 0

4

Sum divisors

sum = 1 + 2 + 3 = 6

5

Compare sum and number

sum (6) == number (6), so number is perfect

i	number % i == 0	sum
1	true	1
2	true	3
3	true	6
4	false	6
5	false	6

Why This Works

Step 1: Find divisors

We find all numbers less than the input that divide it evenly using number % i == 0.

Step 2: Sum divisors

We add these divisors to a sum to get the total of all proper divisors.

Step 3: Compare sum to number

If the sum equals the original number, it means the number is perfect.

Alternative Approaches

Using while loop

csharp

using System;
class Program {
    static void Main() {
        Console.Write("Enter a number: ");
        int number = int.Parse(Console.ReadLine());
        int sum = 0, i = 1;
        while (i < number) {
            if (number % i == 0) sum += i;
            i++;
        }
        Console.WriteLine(sum == number ? $"{number} is a perfect number" : $"{number} is not a perfect number");
    }
}

Uses a while loop instead of for loop; functionally the same but shows different loop style.

Optimized loop up to number/2

csharp

using System;
class Program {
    static void Main() {
        Console.Write("Enter a number: ");
        int number = int.Parse(Console.ReadLine());
        int sum = 0;
        for (int i = 1; i <= number / 2; i++) {
            if (number % i == 0) sum += i;
        }
        Console.WriteLine(sum == number ? $"{number} is a perfect number" : $"{number} is not a perfect number");
    }
}

Checks divisors only up to half the number for better performance.

Complexity: O(n) time, O(1) space

Time Complexity

The program loops from 1 to n (or n/2 in optimized), so time grows linearly with input size.

Space Complexity

Uses only a few variables, so space is constant regardless of input size.

Which Approach is Fastest?

The optimized approach checking up to n/2 is faster than checking up to n, but both are O(n).

Approach	Time	Space	Best For
For loop up to n	O(n)	O(1)	Simple and clear code
While loop up to n	O(n)	O(1)	Alternative loop style
For loop up to n/2	O(n/2) ~ O(n)	O(1)	Better performance for large numbers