\section*{Exercise 1 - 101041}

Form all sets of five one- or two-digit prime numbers such that in each of these sets, each of the digits 1 through 9 appears exactly once!
Let \(M\) be such a set. We give for some digits all one- and two-digit prime numbers in which they are contained as a digit:

\[
2: \quad 2,23,29|4: \quad 41,43,47| 5: \quad 5,53,59|6: \quad 61,67| 8: \quad 83,89
\]

Since the five prime numbers together should have 9 digits, there must be exactly one one-digit and the remaining four two-digit primes.

Case 1: \(2 \in M\). Then the 5 must occur in a two-digit prime number.

Case 1.1: \(53 \in M\). Then \(89 \in M\) must also be true, otherwise the digit 8 would not occur.

Case 1.1.1: \(61 \in M\). Then, to cover the digit 4, \(47 \in M\) must also be true. We get \(M_{1}=\{2,53,89,61,47\}\).

Case 1.1.2: \(67 \in M\). Then, to cover the digit 4, \(41 \in M\) must also be true, so we get \(M_{2}=\{2,53,89,67,41\}\).

Case 1.2: \(59 \in M\). Then, to cover the 8, \(83 \in M\) must also be true.

Case 1.2.1: \(61 \in M\). To cover the 4, \(47 \in M\) must also be true.

We get \(M_{3}=\{2,59,83,61,47\}\).

Case 1.2.2: \(67 \in M\). To cover the 4, \(41 \in M\) must also be true, so \(M_{4}=\{2,59,83,67,41\}\) follows.

Case 2: \(23 \in M\). Then, to cover the digit 8, \(89 \in M\) must also be true. It follows that the 5 can only stand alone, so \(5 \in M\). To cover the digits 4 and 6, there are now again two possibilities, so we get the two sets \(M_{5}=\{23,89,5,61,47\}\) and \(M_{6}=\{23,89,5,67,41\}\).

Case 3: \(29 \in M\). Then, analogous to the second case, \(83 \in M\) and \(5 \in M\) follow. Again, the same two possibilities arise to cover the digits 4 and 6, so we finally get the two sets \(M_{7}=\{29,83,5,61,47\}\) and \(M_{8}=\{29,83,5,67,41\}\).

The case distinction is complete, so there are exactly these eight sets that satisfy the problem statement.