Criteria for divisibility by 9

Question

Prove the following criteria for divisibility by 9:

If $a = \sum\limits_{i=0}^n(c_i10^i)$, where $c_i \in \mathbb{N}$ and $0 \leq c_i < 10$, then $9|a \iff 9|\sum\limits_{i=1}^nC_i$.

Answer 1

$$10^k-1=99\ldots9$$ and $$10^0a_0+10^1a_1+\ldots+10^na_n=\mathbf{(a_0+\ldots+a_n)}+\color{gray}{\left[(10^1-1)a_1+\ldots+(10^n-1)a_n\right]}.$$

Answer 2

In many ways this is really the same thing as RGB's answer, but this can be proven by modular arithmetic. Since $10 \equiv 1 \pmod{9}$, $10^n \equiv 1 \pmod{9}$ and $$ 10^0 a_0+10^1 a_1+\cdots+10^n a_n \equiv a_0+a_1+\cdots+a_n \pmod{9}. $$ So while it's really the same thing, I think seeing it in this language can be helpful.