Decision making and branching - c# program, DOT NET Programming

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Decision Making and Branching - C# Program

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Yuy, Ask queyuyuystion #Minimum 100 words accepted#

Ask queyuyuystion #Minimum 100 words accepted#

james

2/11/2013 7:47:08 AM

This program will help you, try it out

using System;  

class SumOfOdds

{

  public static void Main()

 {

   int x=0, sumodd=0, sumeven=0, sumdiv7 = 0 ,totalno7 = 0, i;    

   for (i=0 ; i<=20 ; i++)

  {

   x = i % 2;

    if (x != 0)

   {

    sumodd = sumodd + i;

   }

    if (x == 0)

   {

    sumeven = sumeven + i;

   }

  }  

   //checking for the sum & no. of numbers divisible by 7    

  x = 0; // resetting the value of ''x''

   for (i=100; i<=200;i++)

  {

   x = i % 7;

    if (x == 0)

   {

    sumdiv7 = sumdiv7 + i;

    totalno7 = totalno7 + 1;

   }   

}     

  Console.WriteLine("Sum of all odd numbers from 1 - 20 = " + sumodd + "\n");

  Console.WriteLine("Sum of all even numbers from 1 - 20 = " + sumeven + "\n");

  Console.WriteLine("Sum of all numbers from 100 - 200, divisible by 7 = " +

sumdiv7 + "\n");

  Console.WriteLine("Total numbers from 100 - 200, divisible by 7  = " + totalno7 + "\n");

   Console.ReadLine();

 }

}

2/11/2013 7:48:14 AM

Hey John i have the ouput of above program:

Output:
Sum of all odd numbers from 1 - 20 = 100
Sum of all even numbers from 1 - 20 = 110
Sum of all numbers from 100 - 200, divisible by 7 = 2107
Total numbers from 100 - 200, divisible by 7  = 14.

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