Euler #4

Next up in Project Euler was a question about palindromic numbers, specifically those that are the product of two three-digit numbers. The text of the problem in its entirety is below.

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009=91×99.

Find the largest palindrome made from the product of two 3-digit numbers.

I set out to write a few lines of brute-force code, but the palindromer() ended up being pretty long. The whole program ran in a flash (way faster than what I wrote for problem #3), though, so I guess I didn't waste too much processing power on this problem.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Problem4
{
    class Program
    {
        static void Main(string[] args)
        {
            int biggest = 10001;
            int product;

            for (int i = 100; i < 1000; i++)
            {
                for (int j = 100; j < 1000; j++)
                {
                    product = i * j;
                    if (palindromeTester(product) && (product > biggest)) { biggest = product; }
                }
            }

            Console.WriteLine(biggest);
        }

        private static bool palindromeTester(int product)
        {
            int palindrome = palindromer(product);
            if (palindrome == product) { return true; }
            else { return false; }
        }

        private static int palindromer(int product)
        {
            int invers = 0;
            int[] bouleverseur = new int[6];

            bouleverseur[5] = (product / 100000);
            product -= bouleverseur[5] * 100000;
            bouleverseur[4] = (product / 10000);
            product -= bouleverseur[4] * 10000;
            bouleverseur[3] = (product / 1000);
            product -= bouleverseur[3] * 1000;
            bouleverseur[2] = (product / 100);
            product -= bouleverseur[2] * 100;
            bouleverseur[1] = (product / 10);
            product -= bouleverseur[1] * 10;
            bouleverseur[0] = (product);

            invers = bouleverseur[5]
                    + bouleverseur[4] * 10
                    + bouleverseur[3] * 100
                    + bouleverseur[2] * 1000
                    + bouleverseur[1] * 10000
                    + bouleverseur[0] * 100000;

            if (bouleverseur[5] == 0) { invers /= 10; }

            return invers;
        }
    }
}