Armstrong Number in C#

Question

I wrote this program in C# to generate Armstrong numbers.I know the addition of extra method is unnecessary but it's something I've just learnt and have been practicing . So the generator only shows number till 9 no matter how large of a limit I give.

using System;

namespace ArmstrongNumberGenerator
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine("Enter the maximum limit for searching");
            int max = Convert.ToInt32(Console.ReadLine());
            for (int i = 1; i <= max; i++)
            {
                if (isArmstrong(i))
                    Console.WriteLine(i);
            }
            Console.ReadLine();
        }

        static bool isArmstrong(int x)
        {
            int temp = x;
            int sum = 0;
            while (x!=0)
            { sum = sum + x % 10;
                x = x / 10;
            }

            if (sum == temp)
                return true;
            else
                return false;
        }
    }
}

I can't see what I am doing wrong.

Answer 1

Explanation for Armstrong-Number:

Sum of the cubes of its digits must equal to the number itself.

For example, 407 is given as input. 4 * 4 * 4 + 0 * 0 * 0 + 7 * 7 * 7 = 407 is an armstrong number.

You need to calculate the sum right:

static bool isArmstrong(int x)
{
    int sum = 0;
    for (int i = x; i > 0; i = i / 10)
    {
        sum = sum + (int)Math.Pow(i % 10, 3.0);
    }

    if (x == sum)
        return true;
    else
        return false;
}

This has following output with your code (except that i incluced 0 which is also an armstrong number):

Enter the maximum limit for searching
9999
0
1
2
3
4
5
6
7
8
9
153
370
371
407
1634
8208
9474