3. On the board, one or several different two-digit natural numbers are written. We will call a digit c on the board good if the sum of those numbers on the board that contain the digit c is equal to the number 71. (Sometimes it may be a "sum" of a single number with the given digit)

a) Which of the digits 0 to 9 can be good?

b) How many digits can be good at the same time?

(Josef Tkadlec)
SOLUTION. a) The following examples of one or two numbers written on the board show that the digits 1, 2, 3, 4, 5, 7 can be good:

$$
\begin{array}{ll}
\text { 1: } & \{71\}, \\
\text { 2: } & \{29,42\}, \\
\text { 3: } & \{32,39\}, \\
\text { 4: } & \{24,47\}, \\
\text { 5: } & \{15,56\}, \\
\text { 7: } & \{71\} .
\end{array}
$$

Let's show that none of the remaining digits $0, 6, 8$, and 9 can ever be good. We will prove this for each of them individually, always speaking about the occurrences of the given digit in the numbers on the board.

- The digit 0 can only appear in the units place. But the sum of such numbers always ends in the digit 0, and not the required digit 1.
- If the digit 6 were only in the units place, the sum of the numbers with the digit 6 would be even, and thus different from 71. So let's assume we have at least one number of the form $6 \sqcup$. For these, however, $6 \sqcup 7171$.

Conclusion of part a). The digits that can be good are precisely 1, 2, 3, 4, 5, or 7.

b) First, we will prove that all the digits $1, 2, 3, 4, 5$, and 7, of which there are 6 in total, cannot be good simultaneously. It is enough to show that the digits 4 and 7 cannot be good at the same time.

Let's examine when the digit 4 is good. Since the number 71 is odd, all the numbers being added cannot have the digit 4 in the units place. At least one number, therefore, has the digit 4 in the tens place, and clearly, there cannot be two such numbers. So we have exactly one number starting with 4 and at least one number ending with 4. However, this number is also unique, as $41+14+24>71$. Therefore, we must have (on the board) exactly two numbers with the digit 4, namely $4 \sqcup$ and $\quad 4$, and their sum is 71. These are certainly the numbers 47 and 24.

Now, let's assume that the digit 7 is good. If it appeared only in the units place, it would have to be in at least 3 numbers to make the sum end in 1, but $17+27+37>71$. Therefore, the good digit 7 must appear in the tens place somewhere - then there is clearly only one number with the digit 7 on the board, namely the number 71.

From the last two paragraphs, it follows what we promised to show: the digits 4 and 7 cannot be good at the same time - on the board, there would have to be the numbers 47, 24, and 71 simultaneously, so the digit 7 would not be good. According to the introduction to part b), this means that the maximum number of good digits on the board is always at most 5 and that the candidates for a set of five simultaneously good digits are only the sets $(1,2,3,4,5)$ and $(1,2,3,5,7)$. To solve the problem, we need to provide an example of numbers on the board with five good digits. For interest, we will provide four such examples, all for the first set $(1,2,3,4,5)$:

$$
\begin{aligned}
& \{10,11,15,16,19\} \cup\{20,24,27\} \cup\{33,38\} \cup\{24,47\} \cup\{15,56\}, \\
& \{10,11,15,17,18\} \cup\{20,24,27\} \cup\{33,38\} \cup\{24,47\} \cup\{15,56\}, \\
& \quad\{15,16,19,21\} \cup\{21,24,26\} \cup\{33,38\} \cup\{24,47\} \cup\{15,56\}, \\
& \quad\{15,17,18,21\} \cup\{21,24,26\} \cup\{33,38\} \cup\{24,47\} \cup\{15,56\} .
\end{aligned}
$$

Conclusion of part b). The maximum number of simultaneously good digits is 5.

Note. First, we will explain why no example of numbers with the set of five good digits $(1,2,3,5,7)$ exists. In the second part of the note, we will at least hint at how to find an example with the set of five good digits $(1,2,3,4,5)$. (Without proof, we leave the fact that in the conclusion of our solution, we have listed all four possible examples.)

It is not difficult to prove that the digit 5 is good only if it appears in two numbers, namely 15 and 56. If the digit 7 is also good, then the numbers 15, 56, and 71 are on the board, so the digit 1 is not good, and the first goal is achieved.

For constructing an example for the set $(1,2,3,4,5)$: We already know that the numbers with the digits 4 or 5 on the board are exactly 24, 47, 15, and 65. By analyzing the numbers with the digit 3, we can find five possible sets: $\{32,39\},\{33,38\},\{34,37\},\{35,36\},\{13,23,35\}$. The last three of these can be immediately ruled out due to the presence of the digits 4 or 5. The first possibility $\{32,39\}$ leads to a problem with the digit 2 - expressing its "deficit" $71-24-32=15$ is not possible. Therefore, the digit 3 must be represented exactly in the numbers 33 and 38. Subsequent analysis of the representation of the digit 2 leads to two possibilities: adding $\{20,27\}$ or $\{21,26\}$ to the number 24. In both cases, the representation of the digit 1 remains to be analyzed, and adding $\{21,26\}$ is simpler - there we are looking for an expression of the "deficit" $71-21-15=35$, for which we can easily find both valid possibilities $16+19$ and $17+18$.

For a complete solution, award 6 points, 3 points for part a) and 3 points for part b). In incomplete solutions, evaluate partial steps as follows.

A1. Determining all six digits $1,2,3,4,5,7$ that can be good, together with providing valid examples - 1 point. Do not award this point if one digit or example is missing, or if an incorrect digit is included.

A2. Determining all four digits $0,6,8,9$ that cannot be good, supported by appropriate justifications - 2 points. A partial 1 point can be awarded if only one of the four digits is omitted.

B1. Proof of the statement that the digits 4 and 7 cannot be good at the same time - 1 point. (No other pair of digits from the set $\{1,2,3,4,5,7\}$ has this negative property.) This point can also be awarded for proving that the digits 1, 5, 7 cannot be good at the same time, or any other statement leading to the same necessary conclusion that there cannot be more than five good digits simultaneously.

B2. Finding one (of the four possible) examples of numbers with five good digits - 2 points.

In total, award $\mathrm{A} 1+\mathrm{A} 2+\mathrm{B} 1+\mathrm{B} 2$ points. If the student does not follow the given scheme but achieves relevant findings, up to 2 points can be awarded (for example, if it is justified which of the digits 4, 5, 6, 7, 8, and 9 can be good and additionally what numbers on the board with each of these good digits must be present). For a mere hypothesis that the maximum number of good numbers on the board is 5, no points should be awarded.