If $a=a'$ mod $n$ and $b=b'$ mod $n$ does $ab=a'b'$ mod $n$? So far this is my working out: 

$n|a-a'$ and $n|b-b'$ 
So $n|(a-a')(b-b')$

Expanding: $n|ab-a'b-ab'+a'b'$
I'm not sure what to do next. I need to show that $n|ab-a'b'$

We can write $a' = a + n k$ for some $k$, and likewise $b' = b + n m$; then
$$a'b' = (a + nk)(b + nm) = ab + n(\text{stuff}) \equiv ab \pmod{n}$$
upon distribution the multiplication.
