Golovanov A.S

In an infinite increasing sequence of natural numbers, each number is divisible by at least one of the numbers 1005 and 1006, but none are divisible by 97. Additionally, any two consecutive numbers differ by no more than \( k \). What is the smallest \( k \) for which this is possible?
Let's denote our sequence as $\left(a_{n}\right)$. It is clear that $a_{1}D$ (while $a_{n} \neq D$ by the condition). However, the largest numbers less than $D$ and divisible by 1005 and 1006 are $D-1005$ and $D-1006$, respectively; therefore, $a_{n} \leq D-1005$. Similarly, $a_{n+1} \geq D+1005$; hence, $a_{n+1}-a_{n} \geq(D+1005)-(D-1005)=2010$. Thus, $k \geq 2010$. For $k=2010$, a sequence of all numbers divisible by 1005 but not divisible by 97 (note that 1005 is not divisible by 97) works.

## Answer

$k=2010$.