M goes into N int( N/M ) times, with a remainder of M%N.
So, M * (1 + int( N/M )) will be the smallest multiple of M greater then N.
--
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In reply to Re: Best way to make sure a number is an even multiple of another?
by TomDLux
in thread Best way to make sure a number is an even multiple of another?
by demerphq
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