How many positive integers $n$ not exceeding 1994 are there such that the fraction

$$
\frac{5 n+13}{11 n+20}
$$

is an integer?
The fraction is always meaningful because the denominator is positive. It can be simplified precisely when the greatest common divisor of $5 n+13$ and $11 n+20$, denoted as $(5 n+13 ; 11 n+20)$, is greater than 1.

Using the identity $(a ; b)=(a ; b+k a)$ multiple times (which is the basic step of the Euclidean algorithm),

\[
(5 n+13 ; 11 n+20)=(5 n+13 ; 11 n+20-2(5 n+13))=(5 n+13 ; n-6)=(5 n+13-5(n-6) ; n-6)=(43 ; n-6).
\]

Since 43 is a prime number, the greatest common divisor can only be 1 or 43. It will be 43 if $n-6$ is divisible by 43. Therefore, the fraction can be simplified for $n$ of the form $43 k+6$. Since $n$ must be between 1 and 1994,

\[
\begin{gathered}
1 \leq 43 k+6=n \leq 1994 \\
-\frac{5}{43} \leq k \leq 46+\frac{10}{43}
\end{gathered}
\]

Since $k$ is an integer, this means that $0 \leq k \leq 46$. There are 47 such $k$ values, so the fraction can be simplified in 47 cases. Braun Gábor (Budapest, Szent István Gymn., II. o.t.)