If all of the digits summed recursively reduce to a 9, then the number is divisible by 9 and also by 3.

If the difference between the sums of alternating sets digits in a number is divisible by 11, then the number itself is divisible by 11.

That’s all I can remember, but yay for math right?

8 points1 day ago* (last edited 1 day ago)There’s also the classic “no three positive integers a, b, and c to satisfy

a**n+b**n=c**nfor values of n greater than 2“ trick but my proof is too large to fit in this comment.