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the last divide() function performs an integer division, the one that should be used inside the first divide() funciton.

the remainder can be calculated like this:

remainder = numerator-integer_divide(n

the remainder of 3/2

remainder = 3-integer_divide(3,2)*2

remainder = 3-1*2 = 1

there may be faster ways to calculate that. i just used the definition of the multiplication and division operations. maybe the processor has a instruction to perform those?