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 $n|ab-a'b'$

In fact, $n^2$ divides your product, not helpful, as it turns out. Your strategy should have been to look at $ab-a'b'$ and show that it’s divisible by $n$. The trick is one you’ve probably seen already in calculus, to add and subtract a clever something. In this case, you write $ab-a'b'=ab-a'b+a'b-a'b'$, then factor the first pair and the second pair separately. Voilà.
