2. Do there exist 2018 pairs of natural numbers $(x ; y)$ that simultaneously satisfy the following conditions: 1) in each pair, $x$ and $y$ do not coincide; 2) in each subsequent pair, the number $x$ is 1 greater than the number $x$ of the previous pair; 3) in each subsequent pair, the number $y$ is 1 greater than the number $y$ of the previous pair; 4) in each pair, $x$ is divisible by $y$?
Solution. Such pairs do exist. It is sufficient to provide an example.

$$
(2018!+1 ; 1),(2018!+2 ; 2),(2018!+3 ; 3), \ldots,(2018!+2018 ; 2018)
$$

Answer. They do exist.