program to get/display digits which are even present in a number.

//program to get/display digits which are even position present in a number.

#include<stdio.h>
#include<conio.h>

void main()
{
   int n,rem,count=0;
   printf("enter a number for 'n'\n");
  scanf("%d",&n);

  while(n!=0)
    {

      rem=n%10;

       count++;

       if(count%2==0)

        {

          printf("%d",rem);

          }

       n=n/10;    

   }

getch();

}

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logics in mind:-

>first we enter a number(for 'n').

->If we have number 123 then its even positioned digits are 2(from both side). It means, first we have to get 3 then 2 and then 1.        

                  ->For this, 

                 ->we have to divide by 10 to get last digit. 

                 ->we also use 'count' to to know position  of digits.

                 ->we divide that count by 2 to get remainder.If count is 2 then we understand that the digit is in even position and if not then that is not.

                 ->Then we display that.

                -> to get second digit, we get first 12 and for this , we use 123/10.It is done to get integer                           part  only.

->We repeat this until the value reaches 0. We use loop for this as shown above in the code.