Every integer is congruent to the sum of its digits mod 9 This is a question from the free Harvard online abstract algebra lectures.  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see this post.
This problem is from assignment 6.  The notes from this lecture can be found here.

Prove that every integer $a$ is congruent to the sum of its decimal digits modulo 9.

Let $s$ be an $(n+1)$-digit integer given by $s=a_n\cdot10^n+\dots+a_2\cdot10^2+a_1\cdot10^1+a_0\cdot10^0$ with $a_i\in\mathbb{Z}$ such that $0\leq a_i\leq9$.  Then
$
\begin{align}
s\,\mathrm{mod}\,9&=\bar{s}\\
&=\overline{a_n\cdot10^n+\cdots+a_2\cdot10^2+a_1\cdot10^1+a_0\cdot10^0}\\
&=\overline{a_n\cdot10^n}+\overline{\cdots}+\overline{a_2\cdot10^2}+\overline{a_1\cdot10^1}+\overline{a_0\cdot10^0}\\
&=\overline{\overline{a_n}\cdot\overline{10^n}}+\overline{\cdots}+\overline{\overline{a_2}\cdot\overline{10^2}}+\overline{\overline{a_1}\cdot\overline{10^1}}+\overline{\overline{a_0}\cdot\overline{10^0}}\\
&=\overline{\overline{a_n}\cdot1}+\overline{\cdots}+\overline{\overline{a_2}\cdot1}+\overline{\overline{a_1}\cdot1}+\overline{\overline{a_0}\cdot1}\\
&=\overline{a_n}+\overline{\cdots}+\overline{a_2}+\overline{a_1}+\overline{a_0}\\
&=a_n\,\mathrm{mod}\,9+\cdots+a_2\,\mathrm{mod}\,9+a_1\,\mathrm{mod}\,9+a_0\,\mathrm{mod}\,9
\end{align}
$
Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.
Thanks.
 A: Your suggestion works if you are satisfied $\overline{a_n\cdot10^n} \equiv \overline{\overline{a_n}\cdot\overline{10^n}} \equiv \overline{\overline{a_n}\cdot1} \equiv \overline{a_n} \mod 9$, which is indeed true. 
Another way is 
$$\begin{align}
s&={a_n\cdot10^n+\cdots+a_2\cdot10^2+a_1\cdot10^1+a_0\cdot10^0}\\
&={a_n\cdot 99\ldots99+\cdots+a_2\cdot 99+a_1\cdot 9 +a_0\cdot 0 }+\sum_{i=0}^n a_i\\
&=9({a_n\cdot 11\ldots11+\cdots+a_2\cdot 11+a_1\cdot 1 +a_0\cdot 0 )}+\sum_{i=0}^n a_i
\end{align}$$
so $s \equiv \sum_{i=0}^n a_i \mod 9.$
A: HINT $\rm\ mod\ 9\!:\ 10\:\equiv\: 1\ \Rightarrow\ f(10)\:\equiv\: f(1)\ $ for all polynomials $\rm\:f(x)\:$ with integer coefficients.
The above applies to radix notation since it clearly has polynomial form in the radix.
Alternatively put $\rm\ x\: =\: 10\ $ in $\rm\ x-1\ |\ f(x)-f(1)\ $ by the Factor Theorem.
A: Just another way.
Clearly $9\vert (10-1)$. Suppose that $9\vert (10^n - 1)$, $n\gt 1$. Then
$$10^{n+1}-1 =10\cdot 10^{n}-1=9\cdot 10^n + (10^n - 1)$$
thus $9\vert (10^{n+1}-1)$. Therefore $9\vert (10^n-1)$ for any $n\in\mathbb{N}$.
Now take an integer
$$a=\sum_{i=0}^n a_i10^{i},$$
where $a_i\in \{0,1,\ldots,9\}$. Then
$$a-\sum_{i=0}^na_i=\sum_{i=0}^n a_i10^{i}-\sum_{i=0}^na_i=\sum_{i=1}^n a_i(10^{i}-1),$$
therefore, by the established above
$$a\equiv \sum_{i=0}^na_i \pmod 9$$
A: I write this answer in keeping with my comment:
You need to prove that $\overline {ab}=\overline{\bar a\bar b}$. There are many ways of looking at this, for instance see, Benjamin Lim's answer, here. I shall write a number theoretic proof here:
Suppose $\bar k$ denotes the remainder obtained when $k$ is divided by$q$, we have, $a=b_1q+\bar a; b=b_2q+\bar b$ . So, you have $$ab=(b_1b_2q+b_1\bar b+b_2\bar a)q+\bar a \bar b$$ and this proves that, $\overline{ab}=\overline{\bar a \bar b}$. 
Further, on the same line, you'll need to prove for the sum and the fact that $10^n \cong 1 (mod~~9)$
There are very minor gaps which you can still fill as a beginner, as others have pointed out.
Sorry for the late edits!!
