On: Faster remainders when the divisor is a constant: ...

[StefanKarpinski]

It's astonishing that this method isn't previously widely known and used. It's so simple once you see it.

[vanderZwan → StefanKarpinski]

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.