IBM Mainframe Forum

This is what I have:



I enter 9780306406157

ISBN NUMBER AFTER DIVIDE: 0000000000000 (note: there are 13 digits)
CHECK DIGIT IS: 10

Supposed to be 978030640615 and 7.
What am I doing wrong?

NVM. SOLVED IT. HAD TO USE BY INSTEAD OF INTO.

It can be done without the need to divide.

A "modulus 10" check-digit is useless. This is just an exercise, it's not how real ISBN's work is it?

Contact [Insert bank of choice] , you may get a job on their Libor team

It is an exercise but it is based on how ISBN-13 numbers are checked in reality. The code of course is longer than what I posted. I can't post the full code until the end of semester. An ISBN-13 number's last digit is a check digit. You add together all odd position digits and multiply by 3. Then add all even position digits. Divide it by 10 and subtract the remainder from 10. The result should equal the check digit.

What's a Libor team? I feel it is some sort of a sarcastic remark.

Hello,

What's a Libor team? I feel it is some sort of a sarcastic remark.

Probably . . . We have lots of this here.

Do you have the check-digit code working now - following the rules you posted?

If not, post where you are stuck and that bit of code and someone should be able to help.

Hey Dick, yep everything works fine now. Thank you and thanks everyone

Hi u written logic correct but where is perform i.e. loop ? it will divide only once right ? if am wrong please correct me..... am just beginner .

Hello and welcome to the forum,

We have lots of beginners here

It is why this forum was started.

The originally posted code was to use as a basis to ask the divide/remainder question. The actual code for the requirement was not posted.

Dependng on how the complete code is written, there may or may not be a loop.