## Task 25/61

A student incorrectly simplifies the fraction $\frac{16}{64}$ by striking out the digit 6 in both the numerator and the denominator. This results in the correct answer $\frac{1}{4}$.

It is to be determined for which fractions with a two-digit numerator and a two-digit denominator this incorrect method also leads to the correct result.
The condition for the feasibility of the method is that the last digit of the numerator matches the first digit of the denominator. Therefore, the fractions can generally be written in the following form:

$$
\frac{10 x+y}{10 y+z}
$$

If the last digit in the numerator and the first digit in the denominator are canceled, the resulting fraction is $\frac{x}{z}$. Both fractions should be equal: $\frac{10 x+y}{10 y+z}=\frac{x}{z}$.

This is an equation in three unknowns, for which only integer solutions between 1 and 9 (both values inclusive) are valid. This is therefore a Diophantine problem. To solve it, proceed as follows: Consider the equation as a proportion and transform it into a product equation: $(10 x+y) z=(10 y+z) x \rightarrow 9 x z=(10 x-z) y$.

The left side of the equation is divisible by 9, so the right side must also be divisible by 9. This is certainly the case if either $y$ or $10 x-z$ is divisible by 9, or if both $y$ and $10 x-z$ are divisible by 3. Thus, the values $3 ; 6 ; 9$ are initially possible for $y$: the case where $10 x-z$ is divisible by 9 will be treated subsequently.

From $y=3$ it follows that $9 x z=3(10 x-z) \rightarrow z=\frac{10}{3+\frac{1}{x}}$. The equation yields an integer $z$ only for $x=3$, giving $z=3$. Thus, $\frac{33}{33}$ is a fraction that satisfies the given condition.

However, we want to consider cases with $x=y=z$ as trivial and only allow solutions where at least $x \neq y$ or $y \neq z$.

Setting $y=6$, we get $z=\frac{20}{3+\frac{2}{x}}$.

For $x=1$, this yields $z=4$, and for $x=2$, it yields $z=5$. This equation has no further integer solutions. Thus, two solutions are found:

$$
\frac{16}{64}=\frac{1}{4} \quad \text { and } \quad \frac{26}{65}=\frac{2}{5}
$$

From $y=9$, we get $z=\frac{10}{1+\frac{1}{x}}$ with the solutions $x=1, z=5$ and $x=4, z=8$ and

$$
\frac{19}{95}=\frac{1}{5} \quad \text { and } \quad \frac{49}{98}=\frac{1}{2}
$$

It remains to consider the case where $10 x-z$ is divisible by 9. For $10 x-z$, only the values 9, 18, 27, 36, 45, 54, 63, 72, 81 are possible; it can be seen that in each case $x=z$. From the equation $9 x z=(10 x-z) y$, it immediately follows that $x=y$, thus yielding only trivial solutions.

There are exactly 4 fractions of the desired type:

$$
\frac{16}{64}=\frac{1}{4} \quad \text { and } \quad \frac{26}{65}=\frac{2}{5} \quad \text { and } \quad \frac{19}{95}=\frac{1}{5} \quad \text { and } \quad \frac{49}{98}=\frac{1}{2}
$$