There are three cards, each with a number from the set $\{1,2, \ldots, 10\}$. These three cards were shuffled and distributed to three people, who recorded the numbers on their respective cards. The cards were then collected and the process was repeated again. After several repetitions, each of the three people summed their records. Knowing that the sums obtained were 13, 15, and 23, what were the numbers on the cards?
Solution

Let $x, y$ and $z$ be the numbers written on the three cards, with $x \leq y \leq z$. At each stage of the drawing process, the sum of the numbers on the three cards is always $x+y+z$. Since $13 + 15 + 23 = 51 = 3 \cdot 17$ and both 3 and 17 are prime numbers, it follows that 3 draws were made and $x+y+z=17$. Analyzing the person who obtained the sum 23, we must have $z \geq 8$, because otherwise the maximum sum would be $7 \cdot 3 = 21$. If $z=10$, in the sums 13 and 15, at most one term $z$ can appear. Since $x+y=17-z=7$, it follows that $(x, y, z)=(3,4,10)$, $(2,5,10)$, or $(1,6,10)$. In the first two cases, it is not possible to obtain the sum 13, and in the last case, it is not possible to obtain the sum 15 with three terms. If $z=8$, then again in the sums 13 and 15, at most one term $z$ can appear. From $3 \cdot 8 > 23 > 8 + 7 + 7$, we can conclude that exactly two terms 8 are used to obtain the sum 23, and thus one of the cards must have the number $23-8-8=7$. Knowing that $x+y=17-z=9$, the only possible solution is $(x, y, z)=(2,7,8)$, but this triplet cannot generate the sums 13 and 15. Finally, the only option left to analyze is $z=9$. In this case, in the sums 13 and 15, exactly one term $z$ is used, and moreover, $x \leq 3$, because otherwise the smallest possible sum with the term $z$ would be $9+4+4=17>13$. On the other hand, since $x+y=17-z=8$, the options left to us are $(x, y, z)=(1,7,9)$, $(2,6,9)$, or $(3,5,9)$. The first cannot generate the sum 13 and the second cannot generate the sum 15. The third option is the only possible one, and an example of draws is illustrated in the following table:

| Person 1 | 3 | 5 | 5 | $=13$ |
| :--- | :--- | :--- | :--- | :--- |
| Person 2 | 9 | 3 | 3 | $=15$ |
| Person 3 | 5 | 9 | 9 | $=23$ |