1. Find all pairs of natural numbers $a, b$ for which the set equality holds

$$
\{a \cdot[a, b], b \cdot(a, b)\}=\{45,180\}
$$

where $(x, y)$ and $[x, y]$ denote the greatest common divisor and the least common multiple of numbers $x$ and $y$, respectively.
1. From the given equality, it follows that the number $b$ is odd (otherwise both numbers on the left would be even), and thus the number $a$ is even (otherwise both numbers on the left would be odd). The equality of sets must therefore be satisfied as follows:

$$
a \cdot[a, b]=180 \quad \text { and } \quad b \cdot(a, b)=45 .
$$

Since the number $a$ divides the number $[a, b]$, the number $180=2^{2} \cdot 3^{2} \cdot 5$ must be divisible by the square of (even) number $a$, so it must be either $a=2$ or $a=6$. In the case of $a=2$ (considering that $b$ is odd), we have

$$
a \cdot[a, b]=2 \cdot[2, b]=2 \cdot 2 b=4 b
$$

which means that the first equality in (1) is satisfied only for $b=45$. Then $b \cdot(a, b)=45 \cdot(2,45)=45$, so the second equality in (1) is also satisfied, and thus the pair $a=2, b=45$ is a solution to the problem.

In the case of $a=6$, similarly, we get

$$
a \cdot[a, b]=6 \cdot[6, b]=6 \cdot 2 \cdot[3, b]=12 \cdot[3, b],
$$

which means that the first equality in (1) is satisfied precisely when $[3, b]=15$. This is satisfied only by the values $b=5$ and $b=15$. However, only the value $b=15$ satisfies the second equality in (1), which is now in the form $b \cdot(6, b)=45$. The second solution to the problem is thus the pair $a=6$, $b=15$, and no other solutions exist.

Answer: The sought pairs are two, namely $a=2, b=45$ and $a=6, b=15$.

Another solution. Let $d=(a, b)$. Then $a=u d$ and $b=v d$, where $u, v$ are coprime natural numbers, so $[a, b]=u v d$. From the equalities

$$
a \cdot[a, b]=u d \cdot u v d=u^{2} v d^{2} \quad \text { and } \quad b \cdot(a, b)=v d \cdot d=v d^{2}
$$

we see that the number $a \cdot[a, b]$ is a $u^{2}$-multiple of the number $b \cdot(a, b)$, so the given set equality can only be satisfied as we wrote the relations (1) in the original solution. These can now be expressed by the equalities

$$
u^{2} v d^{2}=180 \quad \text { and } \quad v d^{2}=45 \text {. }
$$

Therefore, $u^{2}=\frac{180}{45}=4$ or $u=2$. From the equality $v d^{2}=45=3^{2} \cdot 5$ it follows that either $d=1$ (and $v=45$) or $d=3$ (and $v=5$). In the first case, $a=u d=2 \cdot 1=2$ and $b=v d=45 \cdot 1=45$, in the second case, $a=u d=2 \cdot 3=6$ and $b=v d=5 \cdot 3=15$.

Note. Since the given equality immediately implies that both numbers $a, b$ are divisors of the number 180 (even their least common multiple $[a, b]$ is such a divisor), it is possible to provide a complete solution by various other methods based on testing a finite number of specific pairs of numbers $a$ and $b$. We can speed up this process by first determining some necessary conditions that the numbers $a, b$ must satisfy. For example, the refinement of the set equality to a pair of equalities (1) can be (even without using the argument about the parity of the numbers $a, b$) explained by the general observation: the product $a \cdot[a, b]$ is divisible by the product $b \cdot(a, b)$, because their quotient can be written in the form

$$
\frac{a \cdot[a, b]}{b \cdot(a, b)}=\frac{a}{(a, b)} \cdot \frac{[a, b]}{b}
$$

i.e., as a product of two integers.

For a complete solution, award 6 points. If the justification of the equalities (1) is missing in an otherwise complete solution (arguments $(a, b)<[a, b]$ or $(a, b) \mid[a, b]$ alone are insufficient), deduct 1 point. For finding both valid pairs (even by guessing), award 1 point, and the remaining (up to 3) points according to the quality of the provided solution process, especially its systematic nature.