## Task 3 - 231243

Four mathematicians $T, D, S, P$ agree on a guessing game according to the following rules:

$T$ thinks of a triplet $(x, y, z)$ of integers with $1 \leq x \leq y \leq z$ and $x+y+z \leq 10$.

Then he should tell $D$ the number $d=y-x$, $S$ the number $s=x+y+z$, and $P$ the number $p=xyz$, each in such a way that the other two do not learn the value of the communicated number. After that, $D, S$, and $P$ should discuss their information.

Investigate whether there is a triplet $(x, y, z)$ such that in a run of this game (after communicating $d, s$, and $p$), the following conversation can take place:

P: "I cannot determine the triplet $(x, y, z)$ uniquely."

S: "I already knew that before you said it."

P: "Now I can determine the triplet."

D: "I can too."

S: "I can now as well."

If such a triplet exists, determine whether it is uniquely determined by this conversation! If this is the case, provide this triplet!
Given the conditions $x+y+z \leq 10$ and $1 \leq x \leq y \leq z$, for $z_{\text {max }}=8$ there is only the triplet $(z=8: x=1, y=1)$. As $z$ decreases, the combinations shown in the table are added.

Data table

| $\mathrm{z}$ | $\mathrm{y}$ | $\mathrm{x}$ | $\mathrm{y}-\mathrm{x}$ | $\mathrm{z}+\mathrm{y}+\mathrm{x}$ | $\mathrm{zyx}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 8 | 1 | 1 | 0 | 10 | 8 |
| 7 | 1 | 1 | 0 | 9 | 7 |
| 7 | 2 | 1 | 1 | 10 | 14 |
| 6 | 1 | 1 | 0 | 8 | 6 |
| 6 | 2 | 1 | 1 | 9 | 12 |
| 6 | 2 | 2 | 0 | 10 | 24 |
| 6 | 3 | 1 | 2 | 10 | 18 |
| 5 | 1 | 1 | 0 | 7 | 5 |
| 5 | 2 | 1 | 1 | 8 | 10 |
| 5 | 2 | 2 | 0 | 9 | 20 |
| 5 | 3 | 1 | 2 | 9 | 15 |
| 5 | 3 | 2 | 1 | 10 | 30 |
| 5 | 4 | 1 | 3 | 10 | 20 |
| 4 | 1 | 1 | 0 | 6 | 4 |
| 4 | 2 | 1 | 1 | 7 | 8 |
| 4 | 2 | 2 | 0 | 8 | 16 |
| 4 | 3 | 1 | 2 | 8 | 12 |
| 4 | 3 | 2 | 1 | 9 | 24 |
| 4 | 3 | 3 | 0 | 10 | 36 |
| 4 | 4 | 1 | 3 | 9 | 16 |
| 4 | 4 | 2 | 2 | 10 | 32 |
| 3 | 1 | 1 | 0 | 5 | 3 |
| 3 | 2 | 1 | 1 | 6 | 6 |
| 3 | 2 | 2 | 0 | 7 | 12 |
| 3 | 3 | 1 | 2 | 7 | 9 |
| 3 | 3 | 2 | 1 | 8 | 18 |
| 3 | 3 | 3 | 0 | 9 | 27 |
| 2 | 1 | 1 | 0 | 4 | 2 |
| 2 | 2 | 1 | 1 | 5 | 4 |
| 2 | 2 | 2 | 0 | 6 | 8 |
| 1 | 1 | 1 | 0 | 3 | 1 |

Mathematician $D$ can be told one of the differences $\{0,1,2,3\}$ by Mathematician $T$ for $d=y-x$,

Mathematician $S$ with $s=x+y+z$ one of the sums $\{3,4,5,6,7,8,9,10\}$

and Mathematician $P$ with $p=x y z$ one of the products $\{1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,24,27,30,32,36\}$.

If $P$ cannot uniquely determine the triplet, the value of $p(x y z)$ must occur multiple times, and this applies to the products $\{4,6,8,12,18,20,24\}$. If $S$ already knows this before the first statement from $P$, $S$ must have recognized it from $s(x y z)$. Since $s=5$ is the only number that, among those with multiple occurrences of $p(x y z)$, appears only once and shares $p=4$ with $s=6$, $S$ must have received the number $s=6$ from $T$.

Thus, $S$ already knows before the first statement from $P$ that $P$ must have received a product $p=4,6,$ or $8$ from $T$ and therefore cannot uniquely determine the triplet $(x, y, z)$.

After the first statement from $S$, $P$ knows with the received $p=4$ that $S$ must have received $s=6$ from $T$, as $S$ with $s=5$ could not have made the statement (for $s=5$, a unique $p=3$ for $P$ would also have been possible). Thus, the triplet $(x, y, z)=(1,1,4)$ is determinable for $P$.

For Mathematician $D$, it follows that $D$ must have received a $d=0$ from $T$. Now $D$ knows, based on the three statements from $P$ and $S$, that the frequency $n$ of the desired sum must be $2 \leq n(s) \leq 3$, because under different conditions, the three statements would not have been correct.

Thus, only the sums $s=5, s=6$ remain for $D$, and with $d=0$, $s=6$ is determined. With $s=6$, however, two triplets $(x, y, z)=(1,1,4)$ with $p=4$ and $(x, y, z)=(2,2,2)$ with $p=8$ are possible. Since the frequency of $p=8$ is greater than two and thus the triplet would not be determinable for $P$, $D$ can exclude it, and the triplet can be determined.

Mathematician $S$ can also exclude the product $p=8$ from their three possibilities $p=4,6,8$ with the same reasoning. And $D$ must have received a $d=0$, because with a $d=1$, $D$ could not determine a triplet under the given conditions. Thus, $S$ can also determine the triplet.