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 $n|ab-a'b'$

Try it this way:  since $n \mid (a - a')$, we have $a - a' = k_1 n$ for some integer $k_1$; thus $a = a' + k_1n$; likewise $b = b' + k_2 n$ for some $k_2$.  Now when we take $ab$ we get:
$ab = a'b' + a'k_2n + k_1na + k_1k_2n^2$,
so we see that $n \mid (ab - a'b')$. QED.
Hope this helps.  A jovial Yule season to one and all,
and of course,
Fiat Lux!!!
