To find the sum of all the multiples of 3 or 5 below or equal to 1000, first we have to find the sum of all the numbers which are divisible by 3, than sum of numbers which are divisible by 5. Once we are done with this we have to subtract the sum of numbers which are divisible by 3 and 5.

There are two approaches to solve this problem.

1. The first approach is to write a loop and find the numbers which are divisible by 3,5 and add them, after this subtract the numbers which are divisible by 3 and 5.
2. The other approach is to find the count of numbers which are divisible by 3,5 and 15. After this apply the arithmetic progression’s sum formula to find the sum.

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

namespace ConsoleApplication2
{
  class Program
  {
    static void Main(string[] args)
    {
      int sum = 0;
      int range = 1000;
      int Number1 = 3; int Number2 = 5;
      for (int i = 0, j = 0; i <= range; i += Number1, j += Number2)
      {
        sum += i;
        if (j < range && j % Number2 == 0)
        {
          sum += j;
        }
        if (i <= range && i % Number2 * Number1 == 0)
        {
          sum -= i;
        }

      }
      Console.WriteLine("\n");
      Console.WriteLine("Sum of all the multiples of 3 or 5 below or equal to 1000" +
      "using Loop "  + sum);

      int Total_Number1 = range / Number1;
      int Total_Number2 = range / Number2;
      int Total_Number1AndNumber2 = range / (Number1 * Number2);

      int sumTotal_Number1 = Total_Number1 * (Number1 + Total_Number1 * Number1) / 2;
      int sumTotal_Number2 = Total_Number2 * (Number2 + Total_Number2 * Number2) / 2;
      int sumTotal_Number1andNumber2 = Total_Number1AndNumber2 * (Number2 * Number1
          + Total_Number1AndNumber2 * Number2 * Number1) / 2;
      int total = sumTotal_Number1 + sumTotal_Number2 - sumTotal_Number1andNumber2;
      Console.WriteLine("Sum of all the multiples of 3 or 5 below or equal to 1000 " +
                      "using arithmetic progression’s sum formula  " + total);

      Console.ReadLine();
    }
  }
}

Please post your comments if you find a better way to find the sum of numbers which are divisible by 3 or 5 and less than 1000.

Some times it is required to convert data from one base data type to another base data type. The Convert Class converts a base data type to another base data type. Convert class throw a FormatException when the attempt was made to convert a String to other base data type and String value is in not proper format. To Handle FormatException we can use Convert class convert method inside try catch block but there is another better way of doing the same thing is to use TryParse to avoide runtime errors and eliminate the need of Try Catch block.

TryParse method is available for all .Net base data type including Boolean, Char, SByte, Byte, Int16, Int32, Int64, UInt16, UInt32, UInt64, Single, Double, Decimal, DateTime and String. There are two overloaded versions of TryParse method is available for each .Net base data type.

Boolean.Try Parse

Converts specified String value to equivalent Boolean Value. Return True if Conversion is successful otherwise return False.

Byte.TryParse

Converts Specified String value to equivalent Byte Value. Return True if conversion is successful otherwise return False.

TryParse works in the same way for other base type also.