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lang-c
factorfeels a bit awkward here, it gives both too much information (the whole breakdown to prime factors, while just divisibility by 4096 is needed), and too little information (it doesn't help in rounding up)if (remainder != 0) num = num + divisor - remainder; printf "%i ", num, since the other branch is a no-op((x-1)|4095)+1uses bitwise OR to safely round up without overflow, and without an expensive division operation or checking remainders. This of course only works for power-of-2 divisors because it relies on binary integer representations. Does/bin/shsupport$(( ))math at all? I think that's not part of POSIX sh, so your script should be using/bin/bashor/bin/kshor anything other than/bin/sh. Bash supports bitwise-OR with|. If you want to check whether rounding-up happened, check for equality of input vs. output.$(( ))is standard and the standard refers to C operators in the context of arithmetic expansion, so((x-1)|4095)+1should be fine. (It explicitly doesn't require++and--, and the "standalone"(( .. ))construct isn't standard.)