To begin we need to very precisely decide what our question really is. We start by defining some integer, n, with digits arar-1...a1a0. This then means that: n = a0 + 10a1 + ... + 10r-1ar-1 + 10rar.
In order to proceed with our proof we must first prove a simpler fact, that an - bn is always divisible by a - b. We will use induction to prove this lemma.
What this lemma means is that if a ≡ b mod m that an ≡ bn mod m, for any integer n, too. With this fact ready we can now proceed with our proof.
It is clear that 10 ≡ 1 mod 3 (10 - 1 can be perfectly divided by 3), utilising our lemma it is clear that 10k ≡ 1 mod 3. If a ≡ b mod m and c ≡ d mod m then ac ≡ bd mod m (the multiplicative law) and that (a + c) ≡ (b + d) mod m (the additive law), read the proofs here.
0 is obviously divisible by 3, hence ak - ak (the kkt n minus the same digit) is divisible by 3 too - this means that ak ≡ ak mod 3. Utilising the rules of modular arithmetic we have that 10kak ≡ ak mod 3. It then follows from the additive law that n ≡ (ar + ar-1 + ... + a1 + a0) mod 3.
Therefore if 3|n then (ar + ar-1 + ... + a1 + a0) ≡ 0 mod 3, or in English, the sum of the digits of a number that is divisible by 3 will also be divisible by 3. The rule of 3 is proven.
There are also similar laws for division by 9 and 11 but I will leave these proofs up to the reader of this post, you use the exact same techniques and facts as I have used here. Leave any comments on any problems (or successes!) you have with this.