2. Determine all triples of natural numbers $(x, y, z)$ that are solutions to the equation

$$
x y z + x y + y z + z x + x + y + z + 1 = 1985
$$

and $x$ is the smallest of these numbers.
Solution. The given equation is equivalent to the equation

$$
(x+1)(y+1)(z+1)=2 \cdot 3 \cdot 331
$$

Since $x$ is the smallest number, we conclude that $x+1=2$, i.e., $x=1$. Further, $y+1=3, z+1=331$ or $y+1=331, z+1=3$. Thus, we obtain the solutions $x=1, y=2, z=330$ or $x=1, y=330, z=2$.