Yeah, for example to check if a number is divisible by 3, we can use the digit sum and check if the result is divisible by 3, e.g. 153=(1)+(5)+(3)=9. This works

Yeah, for example to check if a number is divisible by 3, we can use the digit sum and check if the result is divisible by 3, e.g. 153=(1)+(5)+(3)=9. This works in base-10, but we can generalize this by converting the dividend X into (3* X+1)-base then doing a digit sum. Let's say we want to check whether 238 is divisible by 17, so our base is 3* 17+1=52, 238 in base-52 is (4)+(30)=34=2* 17. If the convert-to-arbitrary-base operation was particularly cheap to compute, we could convert to base X+1 and check if the last number is 0 (same logic as: is_even = (i & 1) == 0) :)