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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 - 02:25 PM

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 - 02:03 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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