Problem 1052

Answer: No.

Solution. Let $n=10^{1000}$. Denote the original numbers (in order of traversal) as $a_{1}, \ldots, a_{n}$; we will assume that $a_{n+1} = a_{1}$. Let $b_{i} = \operatorname{LCM}(a_{i}, a_{i+1})$. Suppose that the numbers $b_{1}, \ldots, b_{n}$ are $n$ consecutive natural numbers.

Consider the highest power of two $2^{m}$ that divides at least one of the numbers $a_{i}$. Note that none of the numbers $b_{1}, \ldots, b_{n}$ is divisible by $2^{m+1}$. Let, for definiteness, $a_{1} \vdots 2^{m}$; then $b_{1} \vdots 2^{m}$ and $b_{n} \vdots 2^{m}$. Thus, $b_{1} = 2^{m} x$ and $b_{n} = 2^{m} y$ for some odd $x$ and $y$. Without loss of generality, we can assume that $x < y$. Then, since $b_{1}, \ldots, b_{n}$ form $n$ consecutive numbers, among them there must be the number $2^{m}(x+1)$ (since $2^{m} x < 2^{m}(x+1) < 2^{m} y$). But this number is divisible by $2^{m+1}$ (since $x+1$ is even), which is impossible. Contradiction.

Comment. Only the answer - 0 points.

Proved that among the original numbers there are two numbers divisible by $2^{k}$ (the number $k$ is defined in the solution) - 2 points.