3-3. 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 \(n\), \(x_{n}=x\), \(y_{n}=y\), \(z_{n}=z\). Knowing that \(x=1\), find \(y\) and \(z\).
Solve 3. Answer: \(y=z=0\).

The numbers \(x_{n}, y_{n}, z_{n}\) 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 \(x_{n}, y_{n}, z_{n}\) 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.