trying to define even/odd numbers please help

Just figured one (obvious) way to check if the int is even/odd. This is more efficient way, i think (for the computer, that is). This method is based on a simple fact, that all odd integers must end to bit 1, and all even numbers to the bit 0. For non-integer numbers, you still have to use the previous way, i think.

Some examples:
2 => '10'.
3 => '11'.
121 => '111 1001'

If we AND-mask the number 121 with 1 (that is ...0001), heres the result:
111 1001
&&& &&&&
000 0001
=== ====
000 0001

And as we all recall, and operation simply checks if both of the operands are true(1). If both are true, the result is true. 1 AND 1 is true. 1 AND 0 is false etc...

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

namespace EvenOdd
{
    class Program
    {
        static void Main(string[] args)
        {
            int number = -31;
            Console.WriteLine(number + ": " + IsEven(number));

            number = 235578;
            Console.WriteLine(number + ": " + IsEven(number));

            double number2 = 2332.1;
            Console.WriteLine(number2 + ": " + IsEven(number2));

            number2 = -2332.0;
            Console.WriteLine(number2 + ": " + IsEven(number2));
        }

        static bool IsEven(int number)
        {
            //this operation masks the number using bit mask 0x00000001
            //if number's last bit is 1, the result of the bitwise-AND (&) operation is 1.
            //this result is converted to boolean and inverted using !
            return !Convert.ToBoolean(number & 1);
        }

        static bool IsEven(double number)
        {
            bool isEven = false;

            double modulus = number % 2;

            if (modulus == 0)
                isEven = true;
            else if (modulus == 1)
                isEven = false;

            return isEven;
        }
    }
}

thanks so much after little effort it worked.:))

First round your float/double/decimal to nearest integer using Math.Round -method. Then you can calculate the 2-modulus of the integer. If modulus is 0, integer is even. If modulus is 1, integer is odd.

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

namespace EvenOdd
{
   class Program
   {
      static void Main(string[] args)
      {
         double number = 55.15;
         int rounded = Convert.ToInt32(Math.Round(number));
         Console.WriteLine(IsEven(rounded));
      }

      static bool IsEven(int number)
      {
         bool isEven = false;

         int modulus = number % 2;

         if (modulus == 0)
            isEven = true;
         else if (modulus == 1)
            isEven = false;

         return isEven;
      }
   }
}