So I found this on an online platform.
Given three people playing a game of numbers and each one has a positive integer written on their back. None of them know the number written on their own back but they know the numbers written on two other peoples' backs. Given that one of the number is equal to the sum of other two numbers, they are asked, in order, what number is there on their back, each replies they don't know. Now the same question is asked again to the first one and he replies "65", now find the product of the three numbers.
I think this problem doesn't have a proper solution but I am not sure? Is it possible to calculate the product of the three numbers?
Let $x_1,x_2,x_3$ be the numbers on the first, second and third persons' backs respectively. Then $x_1+x_2=x_3\lor x_1+x_3=x_2\lor x_2+x_3=x_1$.
The first says they don't know so it becomes public information that $|x_2-x_3|\in\Bbb Z^+\iff x_2\neq x_3$ and no more information can be deduced.
The second say they don't know and similarly it becomes public that $x_1\neq x_3$. Using $x_2\neq x_3$ it also becomes public that $x_1\neq2x_3$ (the second would've known the answer if otherwise). Nothing else can be deduced.
The third says they don't know and it becomes public that $x_1\neq x_2$. Similarly, it becomes public that $x_1\neq2x_2$ and $2x_1\neq x_2$. Then making use of $x_1\neq2x_3$, it becomes public that $\frac{3}{2}x_1\neq x_2$ and $x_2\neq3x_3$. Nothing else can be deduced.
Finally, the first deduces they have $65$ on their back.
This is possible only if (note $\oplus$ is XOR)
$$\Big((x_1\neq x_2)\land (x_1\neq x_3)\land (x_2\neq x_3)\land (2x_1\neq x_3)\land (x_1\neq 2x_2) \land (2x_1\neq x_2) \land (3x_1\neq 2x_2) \land (x_2\neq3x_3)\Big)\implies\Big(x_1\neq x_2+x_3 \oplus x_1\neq|x_2-x_3|\Big)$$
Note the inequalities involving only $x_2$ and $x_3$ are not helpful and can be ignored.
In other words, the deduction is possible only if $$x_2+x_3\in\left\{\frac{1}{2}x_2,\frac{2}{3}x_2,x_2,2x_2,\frac{1}{2}x_3,x_3\right\}\lor|x_2-x_3|\in\left\{\frac{1}{2}x_2,\frac{2}{3}x_2,x_2,2x_2,\frac{1}{2}x_3,x_3\right\}$$ $\iff$ (going forwards the left term is impossible, since $x_2\neq x_3$ and $x_2+x_3$ is larger than the other members of the set) $$|x_2-x_3|\in\left\{\frac{1}{2}x_2,\frac{2}{3}x_2,\frac{1}{2}x_3\right\}$$ $\iff$ $$x_2\in\left\{\frac{1}{2},\frac{3}{5},\frac{2}{3},\frac{3}{2},2,3\right\}x_3$$ $\iff$ $$x_2\in\left\{\frac{1}{2},\frac{3}{5},\frac{2}{3},\frac{3}{2},2\right\}x_3$$ $\iff$ $$x_2+x_3\in\left\{\frac{3}{2},\frac{8}{5},\frac{5}{3},\frac{5}{2},3\right\}x_3$$
$\iff$ (going forwards $x_1=65=x_2+x_3$ as previously the left term was deemed impossible)
$$65\in\left\{\frac{3}{2},\frac{8}{5},\frac{5}{3},\frac{5}{2},3\right\}x_3$$ $\iff$ $$65\in\left\{\frac{5}{3},\frac{5}{2}\right\}x_3$$ $\iff$ $$(x_1,x_2,x_3)=(65,26,39)\lor(x_1,x_2,x_3)=(65,39,26)$$ $\implies$ $$x_1x_2x_3=65910$$
Going back to the top
If $(x_1,x_2,x_3)=(65,26,39)$:
The first deduces they either have $13$ or $65$ and nothing more.
The second deduces they have $26$ or $104$ and nothing more. The first would've either deduced they could have ($13$ or $65$) or ($65$ or $143$) but nothing more.
The third deduces they have either $39$ or $91$ and nothing more. The first would've either deduced they could have ($13$ or $65$) or ($65$ or $117$) but nothing more. The second would've deduced they could have ($26$ or $104$) or ($26$ or $156$). The second would've then either realised the first would've either deduced they could have (($13$ or $65$) or ($65$ or $143$)) or (($13$ or $65$) or ($65$ or $247$)) but nothing more.
Finally, the first thinks back a bit. If they have $13$, then the third would've deduced that they could only have $13$ or $39$.
The third would've then realised that they couldn't have $13$ as $x_1\neq x_3$ since the second had said they didn't know. The game would've ended there. Therefore the first deduces they must have $65$ and declares so.
If $(x_1,x_2,x_3)=(65,39,26)$:
The first deduces they either have $13$ or $65$ and nothing more.
The second deduces they have $39$ or $91$ and nothing more. The first would've either deduced they could have ($13$ or $65$) or ($65$ or $117$) but nothing more.
The third deduces they have either $26$ or $104$ and nothing more. The first would've either deduced they could have ($13$ or $65$) or ($65$ or $143$) but nothing more. The second would've deduced they could have ($39$ or $91$) or ($39$ or $169$). The second would've then either realised the first would've either deduced they could have (($13$ or $65$) or ($65$ or $117$)) or (($13$ or $65$) or ($65$ or $273$)) but nothing more.
Finally, the first thinks back a bit. If they have $13$, then the third would've deduced that they could only have $26$ or $52$. But if the third had $26$, the second would've deduced that they could only have $13$ or $39$.
The second would've realised that they couldn't have $13$ as $x_2\neq x_3$ since the first had said they didn't know. The game would've ended there. Therefore the third would then deduce that they could only have $52$ and the game would've also ended there. Therefore the first finally deduces they must have $65$ and declares so.