1. Find all triples $a, b, c$ of natural numbers for which the following conditions hold simultaneously:

$$
n(a b, c)=2^{8}, \quad n(b c, a)=2^{9}, \quad n(c a, b)=2^{11}
$$

where $n(x, y)$ denotes the least common multiple of the natural numbers $x$ and $y$.
1. If the numbers $a, b, c$ are solutions to the problem, they are divisors of powers of two, and thus themselves powers of the number $2, a=2^{r}, b=2^{s}, c=2^{t}$, where $r, s, t$ are non-negative integers. From the equality $n(a b, c)=2^{8}$, it follows that the numbers $t, r+s$ are at most 8, with at least one of them equal to 8. Similarly, the numbers $s+t, r$ are at most nine, and at least one of them is equal to nine. Furthermore, one of the numbers $r+t, s$ equals 11, and none of them is greater than 11. However, it cannot be that $s=11$, because $s+t \leqq 9$, so $r+t=11$. It cannot be that $r=9$, because $r+s \leqq 8$. Therefore, $s+t=9$. We then have two possibilities:
1) $t=8$, from which $r=3, s=1, a=2^{3}, b=2, c=2^{8}$,
2) $r+s=8$, from which it follows that $t=6, r=5, s=3$, thus $a=2^{5}, b=2^{3}, c=2^{6}$.

The problem has two solutions.