1.1. The numbers $x, y, z>1$ and $w>0$ are such that $x^{24}=w, y^{40}=w,(x y z)^{12}=w$. Find all $\alpha$ such that $z^{\alpha}=w$.
### 1.1. Answer: 60.

We have $z=(x y z) /(x y)=w^{1 / 12} \cdot w^{-1 / 24} \cdot w^{-1 / 40}=w^{1 / 60}$.