Vanya believes that fractions are "reduced" by crossing out identical digits in the numerator and denominator. Seryozha noticed that sometimes Vanya gets correct equalities, for example, $49 / 98 = 4 / 8$. Find all proper fractions with numerators and denominators consisting of two non-zero digits that can be "reduced" this way.
Let's consider all possible cases of cancellations.

1) $\frac{\overline{\beta a}}{\overline{\overline{b c}}}$. We get $(10 b + a) c = (10 b + c) a$, which simplifies to $b c = b a$. Since $b \neq 0$, it follows that $c = a$. However, by the condition, the fraction is proper. Therefore, there are no solutions.
2) $\frac{\overline{a \bar{b}}}{\overline{\overline{c b}}}$. Similarly to the first case, we find that there are no solutions.
3) $\frac{\overline{\overline{b a}}}{\overline{c \bar{b}}}$. We get $(10 b + a) c = (10 c + b) a$, which simplifies to $9 c(a - b) = b(c - a)$. Since the fraction is proper, $a < b$, so $a - b \geq 1$.

$9 c(a - b) \geq 9 c > 9(c - a) \geq b(c - a)$, which means $9 c(a - b) > b(c - a)$, which is impossible.
4) $\frac{\overline{a \not}}{\overline{b c}}$. Then $(10 a + b) c = (10 b + c) a$, which simplifies to $9 a(b - c) = b(c - a)$. As in the previous case, we note that $b > c > a$. Therefore, $c - a$ cannot equal 9, so $b$ must be a multiple of 3.

If $b = 3$, then $c = 2, a = 1$. But these values do not satisfy the equation.

If $b = 6$, then $c - a = 3$, so $c = 5, a = 2$ or $c = 4, a = 3$. Both options work.

If $b = 9$, then $a(9 - c) = c - a$, which simplifies to $10 a = c(a + 1)$. Since $a$ and $a + 1$ are coprime, 10 must divide $a + 1$. Therefore, $a = 1$ or $a = 4$, from which we get $c = 5$ or $c = 8$.

## Answer

$26 / 65, 16 / 64, 19 / 95, 49 / 98$