In little people school I/we learned that if the sum of a number’s digits are evenly divisible by three, then the number is as well.
This is very cool, and for most people its good enough.
Not for me. I demand proof!
I went digging around the internet, and found squat. So I started in on it, using induction:

Prove it is true for n(1). The first number whose sum is divisible by 3 is 3, which is divisible by 3.

Assume it is true for n. Prove it is true for n+1.
Let X be a number that is evenly divisible by 3. Assume that sum of the digits of X : k+m+n are also evenly divisible by 3. The next number in the series is X+3, whose digits are one of: 
k=k, m=m, n=n+3, for n<7

k=k, m=m+1, n=n7 for n>=7, m!=9

k=k+1, m=m9, n=n7 for n>=7, m=9
If k+m+n is divisible by three, then:
 (k+m+n)+3 is divisible by three, since k+m+n is divisible by 3 and 3 is divisible by 3.
 k+m+1+n7 == k+m+n6 is divisible by three, since k+m+n is divisible by 3 and 6 is divisible by 3.
 k+1+m9+n7 == k+m+n 15 is divisible by three, since k+m+n is divisible by 3 and 15 is divisible by 3.
 For further carryover, we are adding extra 9s, which are divisible by 3, and finally a +1, which makes the 7 a 6, divisible by 3.
QED.
This is great, and I was very proud of myself, but I don’t think it proves what I want it to  I believe I have proved that “if a number is divisible by 3, the sum of its digits is as well” which seems to be a far cry from the converse, what I originally wanted to prove.
Does anyone have an elegant proof for the original statement? Or at least a more elegant proof for the converse, and/or some reasoning why this also means the converse must be true?
Also, I’m interested in proof for the dividebytwo rule (if the one’s digit is evenly divisible by 2, the number is).
Thanks,
Plavacek