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Iwan-Zotow
on Feb 9, 2019
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Faster remainders when the divisor is a constant: ...
see also
https://math.stackexchange.com/questions/1251327/is-there-a-...
acqq
on Feb 10, 2019
[–]
Nice! If I'm not wrong, under many conditions this derivation gives even faster divisibility test then the one presented in the article we comment.
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