this post was submitted on 15 Jul 2025
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You could use a loop to subtract 2 from the number until it equals one or zero
Or literally just look at its binary representation. If the least significant digit is a "1", it's odd, if "0", it's even. Or you can divide by 2 and check for a remainder.
Your method is just spending time grinding away CPU cycles for no reason.
Wasn’t the fun the CPU cycles we wasted along the way?