Answer to POW #7:

Suppose in each case the original number is x. We can write x = 10b + a where a is an integer greater than 0 and strictly less than 10. A theorem suggested by the examples is:

Theorem: If x = 10b + a is divisible by 7, then b - 2a is also divisible by 7.

Proof: (I'm going to use "==" for the usual 3-line congruence symbol often used in modular arithmetic.)
Since x == 0 (mod 7), we have b - 2a == (b - 2a) + 2x (mod 7) = (b - 2a) + 2(10b + a) = 21b, which is clearly == 0 (mod 7).