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.
Perhaps you want $ny \leq x$, otherwise $x = 2$ and $y = 2$ will have no solutions. –  Benjamin Dickman Nov 14 '12 at 4:30
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