\section*{Exercise 3 - 241023}

Determine all natural numbers \(z\) for which the following holds:

If the last digit is removed from the digit representation of \(z\), the resulting digit representation is a divisor of \(z\).
}

According to the problem statement, \(z\) must be at least two digits. Then \(z\) can be represented as \(z=10a+b\), where \(0 \leq b \leq 9\) and \(a=\left[\frac{z}{10}\right]\). After the removal, the remaining number is \(a\).

This means that for \(a\) to be a divisor of \(z\), 

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

must be an integer. This is only possible if either \(b=0\) or \(b\) is a multiple of \(a\). Since in the case \(b \neq 0\), \(b\) can be at most 9, there are only the following possibilities for \(1 \leq a \leq 9\):

\[
\begin{array}{cc|cc|cc|cc}
a & b & a & b & a & b & a & b \\
1 & 1,2, \ldots, 9 & 2 & 2,4,6,8 & 3 & 3,6,9 & 4 & 4,8 \\
5 \ldots 9 & \text { each } b=2a & &
\end{array}
\]

Thus, the following numbers \(z\) are solutions to the problem:

1. all natural numbers ending in 0 with \(z>0\) and
2. \(z \in\{11,12,13,14,15,16,17,18,19,22,24,26,28,33,36,39,44,48,55,66,77,88,99\}\)