Divisibility by $9$

Question

Suppose we have a natural number $N$ with decimal representation $A_kA_{k-1}\ldots A_0$. How do I prove that if the $\sum\limits_{i=0}^kA_i$ is divisible by $9$ then $N$ is divisible by $9$ too?

Answer 1

Hint: $\overline{A_kA_{k-1}\ldots A_1A_0} = 10^kA_k + 10^{k-1}A_{k-1} + \ldots + 10A_1 + A_0$.

$(10^kA_k + 10^{k-1}A_{k-1} + \ldots + 10A_1 + A_0) - (A_k + A_{k-1} + \ldots + A_1 + A_0) = (10^k-1)A_k + (10^{k-1}-1)A_{k-1} + \ldots + (10-1)A_1$ is divisible by $9$.

Answer 2

If you are familiar with modularity you have:

1. If $x \equiv a\mod{m}$ and $y \equiv b\mod{m}$ then $x+y\equiv a+b\mod{m}$.
2. If $x \equiv a\mod{m}$ and $y \equiv b\mod{m}$ then $x\cdot y\equiv a\cdot b\mod{m}$.
3. If $x \equiv 1\mod{m}$ then $x\cdot b\equiv b\mod{m}$.
4. If $x \equiv 1\mod{m}$ then $x^i\equiv 1\mod{m}$.
5. if $x\equiv0\mod m$ then $m$ divides $x$.

So if $A_kA_{k-1}\ldots A_0$ is the decimal representation of $N$, then $$N=\sum_{i=0}^k10^iA_i.$$

First we have that $10\equiv1\mod9$, and therefore (by 4.) $10^i\equiv1\mod9$

Let's evaluate it $\operatorname{mod}9$: $$N=\sum_{i=0}^k10^iA_i \equiv\sum_{i=1}^k1\cdot A_i,\qquad\text{by applying 1. 3. and 4.}$$

I let you continue from here.