Show that if a polynomial $P (x)$ with integer coefficients satisfies $P (x) ≡ 0 \pmod n$ where $n = n_1 × n_2 × \ldots × n_r$ and $n_1, n_2, \ldots , n_r$ are pairwise prime integers then $P (x) ≡ 0 \pmod {n_k}$ for $k= 1, 2, 3, \ldots , r$.
We're given $P(x) \equiv 0 \pmod n$, which means $n$ divides $P(x)$. This means, for some integer $k$
$$ P(x) = k \times n $$
We're also given $n=n_1 \times n_2 \times \ldots \times n_r$, where the $n_k$ are pairwise coprime. This means
$$ \begin{align} P(x) & = k \times (n_1 \times n_2 \times \ldots \times n_k \times \ldots \times n_r) \\ \\ & = k \times \frac{(n_1 \times n_2 \times \ldots \times \cancel{n_k} \times \ldots \times n_r)}{\cancel{n_k}} \times n_k \end{align}$$
And so any $n_k$ divides $P(x)$. That is, for any $k=1,2,\ldots,r$
$$ P(x) \equiv 0 \pmod {n_k} $$
Note that we didn't use the pairwise coprimality of the $n_k$ factors of $n$.