10.5. Given an odd number $n>10$. Find the number of ways to arrange the natural numbers $1,2,3, \ldots, n$ in a circle in some order so that each number is a divisor of the sum of the two adjacent numbers. (Ways that differ by rotation or reflection are considered the same.) (D. Khramov)
Answer. Two ways

Solution. Consider an arbitrary arrangement of numbers from 1 to $n$ that satisfies the conditions. Suppose that two even numbers $x$ and $y$ are adjacent, and the next number is $z$. Since $x+z$ is divisible by $y$, the number $z$ is also even. Continuing this movement around the circle, we get that all numbers in the arrangement are even, which is impossible. Therefore, no two even numbers are adjacent; hence, some two odd numbers are adjacent, and the even and odd numbers alternate.

Note that both neighbors of the number $n$ cannot be even; indeed, otherwise their sum would be divisible by $2n$, i.e., it would be at least $2n$. Therefore, for any odd number less than $n$, either both neighbors are even, or one of the neighbors is the number $n$.

Suppose that the numbers $n$ and $n-2$ are neighbors, and the other neighbor of the number $n-2$ is the number $t$. The number $t+n=(n-2)+(t+2)$ must be divisible by $n-2$, which is only possible if $t=n-4$. But then three odd numbers $n, n-2, n-4$ are adjacent, which, as we have shown, is impossible.

Thus, both neighbors of the odd number $n-2$ are even, and therefore their sum is divisible by $2(n-2)$, i.e., this sum is at least $2(n-2)$; this is only possible if these neighbors are $n-1$ and $n-3$. In particular, the numbers $n-1$ and $n-2$ are neighbors. If the pair $n-1, n-2$ continues with numbers in descending order up to the number 1, we arrive at the circular arrangement $n-1, n-2, n-3, n-4, \ldots, 3, 2, 1, n$, which obviously satisfies the condition.

Now suppose that the pair $n-1, n-2$ continues with numbers in descending order up to the number $d>1$, and after it follows the number $x \neq d-1$. Thus, we have the numbers $n-1, n-2, \ldots, d+1, d, x, y, \ldots$ in a row around the circle. Since $x+(d+1)=(x+1)+d$ is divisible by $d$, then $x+1$ is divisible by $d$, in particular, $x \geq d-1$. But $x$ is different from the numbers $d-1, d, \ldots, n-2, n-1$; hence, the only remaining option is $x=n$.

Let $n=2k+1$. We have obtained that the number $n+1=2k+2$ is divisible by $d$; since $d<10$, the only possible values for $d$ are 1, 2, or 4. If $d=1$, the arrangement is $n, n-1, n-2, n-3, \ldots, 3, 2, 1$, which satisfies the condition. If $d=2$, the arrangement is $2k, 2k-1, 2k-2, \ldots, k+1, 2k+1, k, k-1, k-2, \ldots, 1$, which also satisfies the condition. If $d=4$, the arrangement is $2k, 2k-1, 2k-2, \ldots, k+1, 2k+1, k, k-1, k-2, \ldots, 1$, which also satisfies the condition.

For $n>10$, the arrangement $n, n-1, n-2, n-3, \ldots, 3, 2, 1$ is the only one that satisfies the condition.

Comment. Only the correct answer or the correct answer with the presentation of one of the two required arrangements - 0 points.

Points for the following advancements from different parts a), b), c) are summed.

a) Two examples of arrangements that satisfy the condition are found - 1 point.

b) It is proven that the numbers $n-1$ and $n-2$ are adjacent - 2 points.

If it is only proven that two even numbers do not stand next to each other - 1 point out of these 2.

c) Assuming that $n-1$ and $n-2$ are adjacent, it is proven that the chain uniquely continues to one of the examples $(n-1, n-2, n-3, \ldots, 2, 1, n)$ or $(2k, 2k-1, 2k-2, \ldots, k+1, 2k+1, k, k-1, k-2, \ldots) - 4$ points.

If it is only proven that the continuation has the form $(n-1, n-2, \ldots, d+1, d, n, \ldots)$ for some $d$ - 1 point out of these 4.

If it is proven that the continuation has the form $(n-1, n-2, n-3, \ldots, 2, 1, n)$ or $(2k, 2k-1, 2k-2, \ldots, k+1, 2k+1, k, \ldots) - 3$ points out of these 4.