Then you should return false, unless the remainder is also greater than or equal to the twenty second root of 4194304. Note, that I've only checked up to 4194304 to make sure this works, so if you need bigger numbers, you'll have to validate on your own.

[–] [email protected] 5 points 3 weeks ago (1 children)

i hate to bring this up, but we also need a separate function for negative numbers

[–] [email protected] 1 points 3 weeks ago

You can just bitwise AND those with ...000000001 (for however many bits are in your number). If the result is 0, then the number is even, and if it's 1, then the number is odd. This works for negative numbers because it discards the negative signing bit.