Divisibility by 9

Theorem: A number $$n=a_0+a_110+a_210^2+\ldots+a_n10^n$$ is divisible by 9 if and only if the sum $$a_0+a_1+\ldots+a_n$$ of its digits is divisible by 9.

Direct Proof
If $$n$$ is divisible by 9, then

$$n\equiv0\mod{9}$$ $$\Leftrightarrow a_0+a_110+a_210^2+\ldots+a_n10^n\equiv0\mod{9}$$ $$\Leftrightarrow a_0+a_11+a_21^2+\ldots+a_n1^n\equiv0\mod{9}$$ (since $$10\equiv1\mod{9}$$) $$a_0+a_1+\ldots+a_n\equiv0\mod{9}$$ as desired.

Q.E.D.

Remark
This same argument holds for dvisibility by 3. The proof is exactly the same as the proof above, replacing all instances of 9 by 3.