A fraction has both its numerator and denominator as two-digit integers, and the tens digit of the numerator is equal to the units digit of the denominator. By omitting the common digit of the numerator and the denominator, the quotient of the remaining single-digit numbers is equal to the value of the original fraction. Which is the original fraction?
The task is to find such digits $a, b, c$ that

$$
\frac{10 a+b}{10 c+a}=\frac{b}{c}
$$

Here, $a$ and $c$ are not 0 because the left side is the ratio of two-digit numbers, but $b$ cannot be 0 either, because then the right side would be 0, which would require $a=0$. Removing the fractions,

$$
10 a c+b c=10 b c+a b
$$

which, by rearranging the terms multiplied by 10, can be written as:

$$
10(a-b) c=(a-c) b
$$

Here, if $a-b$ is not 0, then either $a-c$ is equal to 5 or -5, or $b=5$, because the absolute value of both factors is less than 10.

1) If $a-b=0$, then $a-c=0$, i.e., $a=b=c$, and this is clearly a solution to the problem for any positive digit $a$.

smallskip 2) If $a-c=-5, a=c-5$, then from (1) (simplifying by -5)

$$
2(b+5-c) c=b
$$

If $c \leq 5$, then the left side is greater than $b$, and if $c>5$, then the left side is greater than 10, so it cannot be equal to a single digit.

3) If $a-c=5, a=c+5$, then from (1)

$$
2(c+5-b) c=b, \quad(2 c+1) b=2 c(c+5)
$$

From this,

$$
b=c+5-\frac{c+5}{2 c+1}=c+5-\frac{2 c+1+9}{2(2 c+1)}=c+\frac{9}{2}-\frac{9}{2(2 c+1)}
$$

This can only yield an integer if $2 c+1$ is a divisor of 9. This follows for $c=1,4$. In this case, the value of $b$ is 4, or 8, and the value of $a=c+5$ is 6, or 9. The fractions written with these values

$$
\frac{64}{16}=\frac{4}{1}, \quad \frac{98}{49}=\frac{8}{4}
$$

indeed satisfy the requirements of the problem.

4) Finally, if $b=5$, from (1)

$$
2(a-5) c=a-c, \quad(2 c-1) a=9 c
$$

From this,

$$
a=\frac{9 c}{2 c-1}=\frac{9}{2}+\frac{9}{2(2 c-1)}
$$

This is a positive integer if $2 c-1$ is a positive divisor of 9, so $c=1,2,5$. The values of $a$ obtained are 9, 6, 5. The third value pair gives one of the solutions found in 1), and for the other two, the fractions

$$
\frac{95}{19}=\frac{5}{1}, \quad \frac{65}{26}=\frac{5}{2}
$$

also satisfy the requirements of the problem; thus, in addition to 9 trivial solutions, the problem has 4 more solutions.

Remarks. Many contestants expressed one of the digits in terms of the other two based on the requirement, and then checked whether the expression could be a digit for every possible value pair of the latter. We see that the problem can be solved with fewer trials. Many stopped their trial and error after finding one suitable triplet of digits.