If the sum of the digits of $n$ are divisible by 9, then $n$ is divisible by 9; help understanding part of a proof 
Let $n$ be a positive integer such that $n<1000$. If the sum of the digits of $n$ is divisible by 9, then $n$ is divisible by 9. 

I got up to here:
$$100a + 10b + c = n$$
$$a+b+c = 9k,\quad  k \in\mathbb{Z}$$
I didn't know what to do after this, so I consulted the solution
The next step is:
$$100a+10b+c = n = 9k +99a+9b = 9(k +11a+b)$$
I don't get how you can add $99a + 9b$ randomly, can someone please explain this for me?

Simple way to see it: Take the number $N = \sum_{0 \le k \le n} d_k 10^k$ where the $d_k$ are the digits, modulo 9 you have:
$$
N = \sum_{0 \le k \le n} d_k 10^k
  \equiv \sum_{0 \le k \le n} d_k \pmod{9}
$$
since $10 \equiv 1 \pmod{9}$, and so $10^k \equiv 1 \pmod{9}$ for all $k$.
