//program to get/display digits which are even position present in a number.
#include<stdio.h>
#include<conio.h>
#include<conio.h>
void main()
{
int n,rem,count=0;
printf("enter a number for 'n'\n");
scanf("%d",&n);
{
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.
