\section*{Problem 6 - 031216}

It is

\[
\frac{26}{65}=\frac{2 \emptyset}{\phi 5}=\frac{2}{5}
\]

Thus, for this fraction, it is allowed to "cancel" the digit 6. For which fractions with two-digit numerators and denominators is such "cancellation" of any digit of the numerator against a digit of the denominator permitted without changing the represented rational number?
}
Case distinction:
1)

\[
\frac{10 a+c}{10 b+c}=\frac{a}{b} \quad \Rightarrow \quad b c=a c
\]

This means \(c=0\) or \(a=b\). The first implies that all fractions with multiples of 10 in the numerator and denominator can be "canceled" with respect to zero, and the second implies that the numerator and denominator are the same.
2)

\[
\frac{10 c+a}{10 c+b}=\frac{a}{b} \quad \Rightarrow \quad b c=a c
\]

\(c\) cannot be zero, because then the represented numbers would not be two-digit. Therefore, \(a=b\), which leads to an identical numerator and denominator.
3)

\[
\frac{10 a+c}{10 c+b}=\frac{a}{b} \quad \Rightarrow \quad a=\frac{b c}{10 c-9 b}
\]

The possible assignments for \(a, b\), and \(c\) are now to be found by trial and error. The following simplifications apply:

a) The case where all variables are equal has already been addressed.

b) If \(a=x, b=y, c=z\) is a solution, then for all natural numbers \(k\), \(a=k x, b=k y, c=k z\) is also a solution, provided each value is less than 10.

c) For a fixed \(c\), the permissible value range can be further narrowed down:

\[
\begin{aligned}
& 1 \leq a=\frac{b c}{10 c-9 b} \quad \Rightarrow \quad \frac{10 c}{c+9} \leq b \\
& 9 \geq a=\frac{b c}{10 c-9 b} \quad \Rightarrow \quad \frac{10}{9} c \geq b \\
& \begin{array}{|llll|llll}
c & b & \frac{b c}{10 c-9 b} & \text { Solution? } & c & b & \frac{b c}{10 c-9 b} & \text { Solution? } \\
\hline 5 & 4 & 20 / 14 & & 6 & 4 & 24 / 24=1 & \text { yes } \\
6 & 5 & 30 / 15=2 & \text { yes } & 7 & 5 & 35 / 25 & \\
7 & 6 & 42 / 16 & & 8 & 5 & 40 / 35 & \\
8 & 6 & 48 / 26 & & 8 & 7 & 56 / 17 & \\
9 & 5 & 45 / 45=1 & \text { yes } & 9 & 6 & 54 / 36 & \\
9 & 7 & 63 / 27 & & 9 & 8 & 72 / 18=4 & \text { yes }
\end{array}
\end{aligned}
\]
4)

\[
\frac{10 c+b}{10 a+c}=\frac{b}{a}
\]

This case leads to the same relationship as the previous case, only with the numerator and denominator swapped.

The sought fractions are

\[
\frac{16}{64}, \frac{26}{65}, \frac{19}{95}, \frac{49}{98}, \frac{64}{16}, \frac{65}{26}, \frac{95}{19}, \frac{98}{49}
\]

and in addition, all fractions with the same numerator and denominator, as well as all fractions with multiples of 10 in the numerator and denominator.