Another trick: if the last 4 digits of the number are divisible by 16, the full number is divisible by 16. So far as I know, if the last X digits of a number are divisible by 2^X, the entire number is divisible by 2^X.
@James: nice trick! (and I guess the proof arises from 10^x being always exactly divisible by 2^x, so "iff" also applies?)
perhaps less interesting but I guess you could do the same with powers of 5, iff the last X digits of a number are divisible by 5^x, then the entire number is divisible by 5^x...
So many choices
Find a subset that divides the vector into equal halves
Return 'on' or 'off'
Cycling — Critical Power
Divisible by n, Composite Divisors
Divisible by 5
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