4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations

$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=15 \\
\operatorname{LCM}(a ; b ; c)=3^{15} \cdot 5^{18}
\end{array}\right.
$$
Answer: 8568.

Solution. Let $a=3^{\alpha_{1}} \cdot 5^{\alpha_{2}}, b=3^{\beta_{1}} \cdot 5^{\beta_{2}}, c=3^{\gamma_{1}} \cdot 5^{\gamma_{2}}$ (the numbers $a$, $b$, $c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,

$$
\operatorname{LCM}(a ; b ; c)=3^{\max \left(\alpha_{1} ; \beta_{1} ; \gamma_{1}\right)} \cdot 5^{\max \left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)}, \quad \text { GCD }(a ; b ; c)=3^{\min \left(\alpha_{1} ; \beta_{1} ; \gamma_{1}\right)} \cdot 5^{\min \left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)}
$$

Considering the system given in the condition, we obtain the relations

$$
\left\{\begin{array} { l } 
{ \operatorname { max } ( \alpha _ { 1 } ; \beta _ { 1 } ; \gamma _ { 1 } ) = 1 5 , } \\
{ \operatorname { min } ( \alpha _ { 1 } ; \beta _ { 1 } ; \gamma _ { 1 } ) = 1 }
\end{array} \quad \text { and } \quad \left\{\begin{array}{l}
\max \left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)=18 \\
\min \left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)=1
\end{array}\right.\right.
$$

Consider the first system (7). The following sets of numbers $\left(\alpha_{1} ; \beta_{1} ; \gamma_{1}\right)$ are possible:

$(1 ; 1 ; 15)$ - 3 sets (due to different permutations of these numbers);

$(1 ; 15 ; 15)$ - also three sets

$(1 ; k ; 15)$, where $2 \leqslant k \leqslant 14$ - there are 13 different values of $k$, and for each of them 6 permutations - a total of 78 options.

Thus, there are $3+3+6 \cdot 13=84$ ways to choose the triplet of numbers $\left(\alpha_{1} ; \beta_{1} ; \gamma_{1}\right)$. Similarly, we establish that there are $3+3+6 \cdot 16=102$ ways to choose $\left(\alpha_{2} ; \beta_{2} ; \gamma_{2}\right)$, and since one choice is independent of the other, the total number of ways is $84 \cdot 102=8568$.