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C#.NET program to find the sum of odd-positioned digits and even-positioned digits of a given number

By Ashok Nalam on 19 Dec 2013 | Category: C# | Tagged: number .NET 
The program finds the sum of odd-positioned digits and even-positioned digits of a given number using C#.NET example.
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Introduction:

The program finds the sum of odd-positioned digits and even-positioned digits of a given number using C#.NET example.

E.g: If you enter number as 9128456, so even place located numbers are 1,8,5  & Odd place located numbers are 9,2,4,6 and expected output should be 
SumOfEven = 1+8+5=14 
SumOfOdd = 9+2+4+6 =21

Program:

using System;
namespace DotNetMirror
{
    class SumOfOddEvenPositionedDigitsOfNumber
    {
        static void Main()
        {
            Console.Write("Enter number to find sum of Even/Odd positioned digits: ");
            int number = int.Parse(Console.ReadLine());
            int orginalNumber = number;
            int sumEvenPositioned = 0; int sumOddPositioned = 0;
            int tempNumber = 0;
            bool isPointerAtEvenPos = false;
            while (number != 0)
            {
                tempNumber = number % 10; //gives last number
                if (isPointerAtEvenPos)
                {
                    sumEvenPositioned = sumEvenPositioned + tempNumber;
                    isPointerAtEvenPos = false;  //set to odd
                }
                else
                {
                    sumOddPositioned = sumOddPositioned + tempNumber;
                    isPointerAtEvenPos = true; //set to even
                }
                number = number / 10; //removes last number
            }
            Console.WriteLine("Sum of Even positioned numbers in '{0}' is:{1}", orginalNumber, sumEvenPositioned);
            Console.WriteLine("Sum of Odd positioned numbers in '{0}' is:{1}", orginalNumber, sumOddPositioned);
            Console.ReadKey();
        }
    }
}

Output:

sum of odd-positioned digits and even-positioned digits of a given number output
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