6. Find all triples of natural numbers $a, b, c$ for which the set equality

$$
\{(a, b),(a, c),(b, c),[a, b],[a, c],[b, c]\}=\{2,3,5,60,90,180\},
$$

holds, where $(x, y)$ and $[x, y]$ denote the greatest common divisor and the least common multiple of numbers $x$ and $y$, respectively.
SOLUTION. We factorize the elements of the given set $M$:

$$
\mathrm{M}=\left\{2,3,5,2^{2} \cdot 3 \cdot 5,2 \cdot 3^{2} \cdot 5,2^{2} \cdot 3^{2} \cdot 5\right\}
$$

From this, it follows that in the factorization of the sought numbers $a, b, c$, only the prime numbers 2, 3, and 5 appear. Each of them is a prime factor of exactly two of the numbers $a, b, c$: if it were a prime factor of only one of them, it would be missing in the factorization of two of the greatest common divisors and one of the least common multiples, i.e., in three numbers from $M$; if, on the other hand, it were a prime factor of all three numbers $a, b, c$, it would not be missing in the factorization of any number from $M$. Moreover, we see that in the factorization of each of the numbers $a, b, c$, the prime number 5 appears at most once.

According to the above findings, we can arrange the numbers $a, b, c$ such that the factorizations of the numbers $a, b$ each contain one instance of the prime number 5 (while $(c, 5)=1$) and that $(a, 2)=2$ (as we know, at least one of the numbers $a, b$ must be even). The number 5 from the set $\mathrm{M}$ is then necessarily equal to $(a, b)$, so $(b, 2)=1$, and therefore $(b, 3)=3$ (otherwise $(b, c)=1$), from which, considering $(a, b)=5$, it follows that $(a, 3)=1$. We thus have $a=5 \cdot 2^{s}$ and $b=5 \cdot 3^{t}$ for suitable natural numbers $s$ and $t$.

From the equality $[a, b]=2^{s} \cdot 3^{t} \cdot 5$, it follows that one of the following three cases occurs.

(1) $2^{s} \cdot 3^{t} \cdot 5=60=2^{2} \cdot 3^{1} \cdot 5$. We see that $s=2$ and $t=1$, i.e., $a=20$ and $b=15$.

It is easy to determine that the third number is $c=18$.

(2) $2^{s} \cdot 3^{t} \cdot 5=90=2^{1} \cdot 3^{2} \cdot 5$. In this case, $a=10, b=45$, and $c=12$.

(3) $2^{s} \cdot 3^{t} \cdot 5=180=2^{2} \cdot 3^{2} \cdot 5$. Now $a=20, b=45$, and $c=6$.

Answer: The sought numbers $a, b, c$ form one of the sets $\{10,45,12\},\{20,15,18\}$, and $\{20,45,6\}$.