Testing if a list of integer is odd or even

        #region even and odd numbers
        for (int x = 0; x <= 50; x = x + 2)
        {

            int y = 1;
            y = y + x;
            if (y < 50)
            {
                Console.WriteLine("Odd number is #{" + x + "} : even number is #{" + y + "} order by Asc");
                Console.ReadKey();
            }
            else
            {
                Console.WriteLine("Odd number is #{" + x + "} : even number is #{0} order by Asc");
                Console.ReadKey();
            }

        }

        //order by desc

        for (int z = 50; z >= 0; z = z - 2)
        {
            int w = z;
            w = w - 1;
            if (w > 0)
            {
                Console.WriteLine("odd number is {" + z + "} : even number is {" + w + "} order by desc");
                Console.ReadKey();
            }
            else
            {
                Console.WriteLine("odd number is {" + z + "} : even number is {0} order by desc");
                Console.ReadKey();
            }
        }

You could try using Linq to project the list:

var output = lst.Select(x => x % 2 == 0).ToList();

This will return a new list of bools such that {1, 2, 3, 4, 5} will map to {false, true, false, true, false}.

There's at least 7 different ways to test if a number is odd or even. But, if you read through these benchmarks, you'll find that as TGH mentioned above, the modulus operation is the fastest:

if (x % 2 == 0)
               //even number
        else
               //odd number

Here are a few other methods (from the website) :

//bitwise operation
if ((x & 1) == 0)
       //even number
else
      //odd number

//bit shifting
if (((x >> 1) << 1) == x)
       //even number
else
       //odd number

//using native library
System.Math.DivRem((long)x, (long)2, out outvalue);
if ( outvalue == 0)
       //even number
else
       //odd number

Just use the modulus

loop through the list and run the following on each item

if(num % 2 == 0)
{
  //is even
}
else
{
  //is odd
}

Alternatively if you want to know if all are even you can do something like this:

bool allAreEven = lst.All(x => x % 2 == 0);