9.5. How many solutions in natural numbers $x, y$ does the system of equations have

$$
\left\{\begin{array}{l}
\text { GCD }(x, y)=20! \\
\text { LCM }(x, y)=30!
\end{array} \quad(\text { where } n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n) ?\right.
$$
Answer: $2^{8}$. Solution. If for the given two numbers $x, y$ the set of their prime divisors is denoted as $p_{1}, p_{2}, \cdots, p_{k}$ and we write $x=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}, y=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}}$ (where $\alpha_{i}, \beta_{i}$ are non-negative integers), then

$$
\begin{aligned}
& \operatorname{GCD}(x, y)=p_{1}^{\min \left(\alpha_{1}, \beta_{1}\right)} \cdot p_{2}^{\min \left(\alpha_{2}, \beta_{2}\right)} \cdots \cdots p_{k}^{\min \left(\alpha_{k}, \beta_{k}\right)} \\
& \operatorname{LCM}(x, y)=p_{1}^{\max \left(\alpha_{1}, \beta_{1}\right)} \cdot p_{2}^{\max \left(\alpha_{2}, \beta_{2}\right)} \cdots \cdots p_{k}^{\max \left(\alpha_{k}, \beta_{k}\right)}
\end{aligned}
$$

The problem is to find $x, y$ for given $A, B$ from the equations GCD $(x, y)=A$, LCM $(x, y)=B$. If the factorizations of $A$ and $B$ are $A=p_{1}^{n_{1}} p_{2}^{n_{2}} \cdots p_{k}^{n_{k}}, B=p_{1}^{m_{1}} p_{2}^{m_{2}} \cdots p_{k}^{m_{k}}$ and $n_{1}=m_{1}$, then $\alpha_{1}=\beta_{1}$. If $n_{1}n_{i}$, specifically 2,3, 5,7,11,13,23,29 (all prime numbers less than 30 except 17 and 19). Therefore, the number of different ways to choose ordered sets $\alpha_{i}, \beta_{i}$ is $2^{8}=256$.