Delhi odd even please chk code

#1

please check my code TEST CASE is FAILING
sample output is correct

my code ; https://ide.codingblocks.com/s/70709

#2

Hii Nipun,
Your logic is wrong as in this question you have to find sum of digits at even place in a number and sum of digits at odd places in a number and you need to check is sum of digits at even places is divisible by 4 or sum of digits at odd places is divisible by 3. If yes, print yes else print no.

#3

how to find the place of the digits in number

#4

Hi Nipun

You need to find the digit at each individual place in the number and add it to the respective even / odd sum. And then check for possible conditions.

This is quite an easy question and shouldn’t be much of an issue.

Hint: For number n = 93, we can get the value of 3 by

int rem = n % 10; // rem = 3
n = n / 10; // n = 9

Similarly, we can 9 by

int rem2 = n % 10; // rem2 = 9

Try to understand from this. Hope this helps.

#5

Hi Nipun

As you are not responding to this thread, I am marking your doubt as Resolved for now. Re-open it if required.

Please mark your doubts as resolved in your course’s “ Ask Doubt ” section, when your doubt is resolved.