My problem is such:
On a circle there are $9$ distinct positive integers aranced in such a way that the product of two non-adjacent numbers in the circle is a multiple of $n$ and the product of any two adjacent numbers in the circle is not a multiple of $n$. Here, $n$ is a fixed positive integer. What is the smallest possible value for $n$?
I have found a solution if someone is willing to compare answers with mine. My answer came out to be 485100. Can someone please verify this?
Denote the integers $n_1, \dots, n_9$.
For each $i = 1, \dots, 9$, the product $n_i n_{i+1}$ is not a multiple of $n$ (where the indices should be read modulo nine, i.e., $n_{10} = n_1$). Therefor, there exists a prime $p_i$ such that $p_i$ divides $n$ but $p_i$ does not divide $n_i$ or $n_{i+1}$.
If $p_i$ does not divide $n_j$ for some $j \neq i, i+1$, then neither $n_jn_i$ nor $n_jn_{i+1}$ is a multiple of $n$. Since $n_j$ is non adjacent to either $n_i$ or $n_{i+1}$, this is a contradiction. Hence, $n_i$ and $n_{i+1}$ are the only two integers not divisible by $p_i$.
Since there are nine pairs of adjacent integers, $n$ must have at least nine prime factors $p_1, \dots, p_9$.
On the other hand, $$n_i = p_1 \cdots \hat p_{i-1} \hat p_i \cdots p_9,$$ solves the problem with $$n = p_1 \cdots p_9.$$
The smallest such integer is $$n = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 = 223 092 870$$