We took three numbers $x, y, z$. We calculated the absolute values of the pairwise differences $x_{1}=|x-y|, y_{1}=|y-z|, z_{1}=|z-x|$. In the same way, from the numbers $x_{1}, y_{1}, z_{1}$, we constructed the numbers $x_{2}, y_{2}, z_{2}$, and so on. It turned out that for some $n$, $x_{\mathrm{n}}=x, y_{\mathrm{n}}=y, z_{\mathrm{n}}=z$. Knowing that $x=1$, find $y$ and $z$.
Answer: $y=z=0$. The numbers $x_{n}, y_{\mathrm{n}}, z_{\mathrm{n}}$ are non-negative, so the numbers $x, y, z$ are also non-negative. If all the numbers $x, y, z$ were positive, then the largest of the numbers $x_{1}, y_{1}, z_{1}$ would be strictly less than the largest of the numbers $x, y$, $z$, and then the largest of the numbers $x_{\mathrm{n}}, y_{\mathrm{n}}, z_{\mathrm{n}}$ would be strictly less than the largest of the numbers $x, y, z$. Therefore, among the numbers $x, y, z$ there is a 0. Similarly, it is proved that among the numbers $x_{1}, y_{1}, z_{1}$ there is a 0 (for $n=1$ there is nothing to prove, because then $x_{1}=x, y_{1}=y, z_{1}=z$). This means that two of the numbers $x, y, z$ are equal. In the end, we get that the unordered set of numbers $x, y, z$ can be either $0,0,1$ or $0,1,1$. Obviously, the second set does not have the required property.