2A. Determine how many ordered pairs of natural numbers $(x, y)$ exist for which $\operatorname{LCM}(x, y)=6$!
Solution. Since $6!=2^{4} \cdot 3^{2} \cdot 5^{1}$ and $x=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}}, y=2^{b_{1}} \cdot 3^{b_{2}} \cdot 5^{b_{3}}$, from the condition of the problem it follows that $\max \left\{a_{1}, b_{1}\right\}=4, \max \left\{a_{2}, b_{2}\right\}=2, \max \left\{a_{3}, b_{3}\right\}=1$. According to this, the pairs $\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right)$ and $\left(a_{3}, b_{3}\right)$ can be chosen in 9, 5, and 3 ways (by direct verification). It follows that there are $9 \cdot 5 \cdot 3=135$ ordered pairs for which the condition of the problem is satisfied.