There's a pretty well-known test for divisibility by seven that came up over coffee today. Namely, to check if a number written in base 10 is divisible by 7, remove the last digit; then subtract twice that digit from the "rest" of the number. The input is divisible by 7 if and only if the output is; since this test makes numbers smaller, that's fine.
For example, 4728402 is divisible by 7. Why? Remove the last digit (2) and subtract its double (4) from the "rest" of the number, giving 472840 - 4 = 472836. Repeating, we get:
47283 - 12 = 47271
4727 - 2 = 4725
472 - 10 = 462
46 - 4 = 42
4 - 4 = 0
which is divisible by 7. (The number isn't special in any way; I just pressed a bunch of number keys and multiplied by 7 to create it.)
But I'd never seen a proof that this works. Here's a quick one. We want to show that 10k + m is divisible by 7 if and only if k - 2m is.
Say 10k + m is divisible by 7. Then 3k+m is, so 10k+m - 3(3k+m) = k-2m is.
Conversely, say k-2m is divisibile by 7. Then 3(k-2m) + 7(k+m) = 10k+m is.