Problem 786

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.