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Mod Function ( A Mod B) How To


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#1 adrios

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Posted 01 September 2012 - 03:25 PM

Hi!
I' m new here :)

I' m casio fx-7400G owner, and was wondering about creating a program to do "a mod b" in a easy way, instead of using the subtraction one.

is it a common a question?
can you help me?

hope this is the right section, thank you

#2 MicroPro

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Posted 25 September 2012 - 03:03 PM

Welcome to UCF, adrios! Introduce yourself here.

What do you want to do? a mod b can be calculated by b * (a/b - int(a/b)). ("int" is the integer part function)
So there is no need to subtractions.

I may not have understood you correctly. Please explain more?




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