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I have a question. For $x,y,n \in\Bbb N$, $y$ a power of two, being given $x$ and $y$ is there a faster way to mentally calculate $ny$ where $ny ≤ x< (n+1)y-1$ other than $\lfloor x \div y \rfloor \times y$? Thanks.

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Perhaps you want $ny \leq x$, otherwise $x = 2$ and $y = 2$ will have no solutions. –  Benjamin Dickman Nov 14 '12 at 4:30
    
@B.D yes, thank you –  octavian Nov 14 '12 at 4:42
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All I can think of is the usual $2^{10} \approx 1,000$ and its powers can get you $n$ with less calculation (as long as it isn't too close). –  Ross Millikan Nov 14 '12 at 4:56
    
@RossMillikan I figured out one thing you could do would be x-x%y, that also gives you ny –  octavian Nov 15 '12 at 8:40

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