Problem 1129

First observe that for each index $i \in \{1, 2, \cdots, 9\}$,
$$D_i = \prod_{1 \le j \le 9,\ j \ne i} |d_i - d_j|$$
is a positive number. Let $N = \max\{D_1 - d_1, D_2 - d_2, \cdots, D_9 - d_9\}$. We will show below that $N$ satisfies the requirement of the problem.

Suppose there exists an integer $x \ge N$ such that all prime factors of $P(x)$ are less than 20. For all $1 \le i \le 9$, consider the reduced fraction of $(x+d_i)/D_i$, denoted as $A_i/B_i$. Note that $A_i|P(x)$, and since $x+d_i \ge (D_i-d_i+1)-d_i > D_i$, $A_i > 1$, so there must exist a prime $p_i$ less than 20 such that $p_i|A_i$. Moreover, since there are only 8 primes less than 20, by the pigeonhole principle there exist $1 \le i < j \le 9$ such that $p_i = p_j = p$. Hence there exist positive integers $\alpha_i, \alpha_j$, and positive integers $q_i, q_j$ coprime to $p$, such that $x+d_i = p^{\alpha_i}q_i$ and $x+d_j = p^{\alpha_j}q_j$. Without loss of generality, assume $\alpha_i < \alpha_j$.

But at the same time, note that $p^{\alpha_i}|A_i|x+d_i$ and $p^{\alpha_i}|p^{\alpha_j}|A_j|x+d_j$, so $p^{\alpha_i}|d_i-d_j|D_i$, hence
$$p|A_i \frac{x+d_i}{D_i} \frac{p^{\alpha_i}q_i}{p_i^{\alpha_i}} = q_i,$$
but $q_i$ is coprime to $p$, a contradiction! Therefore $N$ satisfies the requirement of the problem.