\section*{Exercise 3 - 071013}

A mathematician has lost the key to the compartment of a luggage machine. However, he still remembered the number of the compartment, which was a three-digit number divisible by 13, and that the middle digit was the arithmetic mean of the other two digits. The compartment could be quickly identified because only a few numbers have these properties.

List all such numbers!
}
According to the task, we have:

\[
\begin{gathered}
100 a + 10 b + c = 13 n \quad (a, b, c, n \text{ integers}) \\
00
\end{gathered}
\]

From this, it follows that:

\[
c = 2 b - a
\]

and \(99 a + 12 b = 13 n\)

\[
b = \frac{13 n - 99 a}{12} = n - 8 a + \frac{n - 3 a}{12}
\]

Thus, \(n - 3 a = 12 m\) (where \(m\) is an integer) must hold, from which we get

\[
n = 12 m + 3 a
\]

From (3) and (4), we obtain

\[
b = 13 m - 5 a
\]

From (2) and (5), we get

\[
c = 26 m - 11 a
\]

and from (5) and (6), \(16 a + b + c = 39 m\), which, considering (1), gives \(17 \leq 39 m \leq 162\). This results in

\[
1 \leq m \leq 4
\]

Since \(m > 0\) and must be an integer, from (6) and (1) we get

\[
11 a \leq 26 m \leq 11 a + 9
\]

Using (7) and (8), we determine from (5) and (2) that the only possible numbers are 234, 468, 741, and 975, as each value of \(m\) in (7) provides exactly one value for \(a\) in (8). Since each of these four numbers satisfies all conditions of the task, they are also solutions.