Example 6. In the six-digit number $31 x y 13$ !, $x 、 y$ are both greater than 5, and this number is divisible by 13, find the four-digit number $\overline{1 x y 9}$.
$$
\begin{array}{l}
\text { Solution: Let } n=31 x y 13 \\
=3 \cdot 10^{5}+10^{4}+x \cdot 10^{3}+y \cdot 10^{2} \\
+13 \\
=310013+10^{3} x+10^{2} y, \\
\end{array}
$$

where $14,6 \leqslant x, y \leqslant 9$.
Since the remainders of $310013,10^{3}, 10^{2}$ when divided by 13 are $2,-1,-4$ respectively, we can set $310013=13 k_{1} + 2$, $10^{2} x=13 k_{2}-x, 10^{2} y=13 k_{3}-4 y$. Let $k=k_{1}+k_{2}+k_{3}\left(k_{1}, k_{2}, k_{3} \in \mathbb{N}\right)$, then
$$
n=13 k+2-x-4 y.
$$
$\because 13|n, \quad \therefore 13| 2-x-4 y$.
Let $2-x-4 y=13 t(t \in \mathbb{Z})$.
From $6 \leqslant x, y \leqslant 9$ we get
$$
-43 \leqslant 13 t \leqslant-28.
$$

Therefore: $t=-3$. Substituting $-x-4 y=13 t$, we get
$$
x + 4 y=41, \quad \text { and } y=10+\frac{1-x}{4}.
$$

Thus, $x=9, y=8$. The required four-digit number is 1989.
This problem mainly uses the divisibility of numbers, and the method is also quite special.

For such problems, it is not necessary to express integers as polynomials, but since this is not within the scope of this article, it will not be discussed further here.
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