If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$ The question is If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$.
My attempt is that $b \equiv 0 \pmod a$ can be written $a\mid b-0 = a\mid b$
and the same with $c \equiv 0 \pmod b$ can be written $b\mid c-0 = b\mid c$
then the $c \equiv 0 \pmod a$ can also be written $a\mid c-0 = a\mid c$.
Now all I have to show is that $a\mid b$ and $b\mid c$ then $a\mid c$ and to show $a\mid c$ I can write
$c= ak$ for some integer $k$.
Now I start my proof 
let $a\mid b$ be $b=a\cdot r$ for some integer $r$ (1)
let $b\mid c$ be $c=b\cdot s$ for some integer $s$ (2)
Sub (1) into (2) and get
$c=(ar)\cdot s$
$c=a\cdot(r\cdot s)$
so $c=a\cdot k$
therefore the statement $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$ is true.
I'm wondering if this is a good/correct proof?
any input would be appreciated.

You did not define what $k$ is, so no, your proof is not correct. You simply state

So $c=ak$.

What is $k$? How did you prove that $c=ak$?
