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Re:Faster remainders when the divisor is a constant: beating... by StefanKarpinski on Friday February 08, 2019 @04:11PM (#19118702)
It's astonishing that this method isn't previously widely known and used. It's so simple once you see it.
Re:Faster remainders when the divisor is a constant: beating... by vanderZwan on Saturday February 09, 2019 @02:35AM (#19121208)
I wouldn't be surprised if some people had already thought of it while coming up with a fast quotient method, but didn't bother to do anything with it because they did not need a remainder.