I can prove the 9 rule, although it's hard to type out.
Let n be a positive integer such that the sum of the digits is divisible by 9. We can write the digits of n as dk, dk-1, dk-2, ..., d1, d0. (Supposed to be subscripts) So d0+d1+d2+...+dk=9m for some integer m. We can write out the decimal expansion of n as 10^k*dk+10^(k-1)*dk-1+...+10d1+d0. Separate this sum into [(999...9)dk+(999...9)dk-1+...99d2+9d1]+[d0+d1+d2+...+dk]. The first sum is divisible by 9, so we can equate it to 9s. The second sum is 9m (from above). So we have n=9s+9m=9(s+m). Hence, 9 divides n.
I hope this helps. You can prove a lot of the divisibility rules this way. I teach a discrete math course for gifted hs students. A great resource for number theory is the Epp discrete math book.
