Does this wrong cancellation of $B$ work for $\overline{AB}/ \overline{BC}=A/C$? My teacher says that wrong cancellation of $B$ for the fraction$$\frac{\overline{AB}}{\overline{BC}}=\frac{A}{C}$$ will work for some  numbers. I see some trivial cases when $A=B=C$, but are there more of them?
$A$, $B$ and $C$ are some digits.

Interesting question! I've solved this before! Let $\overline{AB}=10a+b$, $\overline{BC}=10b+c$, we get$$\frac{10a+b}{10b+c}=\frac{a}{c}$$ and $$b(10a-c)=9ac$$Note that $a$, $b$ and $c$ must be nonzero digits. $c=a$ is our trivial case, which leads to $a=b=c$. For $a \neq c$ let $d=c-a$. Its clear that $d$ cannot take $\pm9$. Equation becomes $$b(9a-d)=9a(a+d) \\ 9a^2+9a(d-b)+bd=0$$It's immediate that $9|bd$ and since $d\neq \pm9$ only possible values for $b$ are $3$, $6$ and $9$.
For $b=3$ $$c=\frac{10a}{3a+1}$$and we must have $3a+1|10(3a+1)-3(10a)=10$, which gives $a=3$ only and this is one of trivial solutions.
For $b=6$ $$c=\frac{20a}{3a+2}$$and we must have $3a+2|20(3a+2)-3(20a)=40$, which gives $a=1, 2, 6$ and leads to $(a, b, c)=(1, 6, 4), (2, 6, 5)$ and trivial solution $(6, 6, 6)$.
For $b=9$ $$c=\frac{10a}{a+1}$$and we must have $a+1|10(a+1)-10a=10$, which gives $a=1, 4, 9$ and leads to $(a, b, c)=(1, 9, 5), (4, 9, 8)$ and trivial solution $(9, 9, 9)$.
We are done. 
