Show that if $ab \equiv ac$ mod $n$ and $d=(a,n)$, then $b \equiv c$ mod $\frac{n}{d}$ What I know so far: 
We know by the definition of congruence that $n$ divides $ab-ac$. So, there exists an integer $k$ such that $a(b-c)=kn$, and since $d=(a,n)$ we know that $a=ds$ and $n=dt$ from some integers $s$ and $t$.  Then substituting for $a$ and $n$, we see that $ds(b-c)=k(dt)$.

Since $\gcd(a,n)=d$, we get
$$\gcd\bigg(\frac{a}{d},\frac{n}{d}\bigg)=1.$$ Write 
$$r=\frac{a}{d}\quad\text{and}\quad s=\frac{n}{d}.$$ Then $$\gcd(r,s)=1\quad\text{and}\quad \frac{a}{n}=\frac{r}{s}.$$ Also, we get
$$n\big|(ab-ac).$$ Hence,
$$\frac{ab-ac}{n}\in\Bbb Z.$$ But,
$$\begin{align}
\frac{ab-ac}{n}&=\frac{a(b-c)}{n}\\
&=\frac{r(b-c)}{s}.
\end{align}$$
Hence,
$$\frac{r(b-c)}{s}\in\Bbb Z.$$
Thus,
$$s\big|r(b-c).$$ 
Since $\gcd(r,s)=1$, using the Euclid's Lemma, we get
$$s\big|(b-c).$$
Hence, $$b \equiv c \mod s.$$
Thus, $$b \equiv c \mod{\frac{n}{d}}.$$
