$12 \cdot 60$ In a right-angled triangle with side lengths all being positive integers, one of the legs is 1575 less than the hypotenuse, and the other leg is less than 1991. Find the length of the hypotenuse of this right-angled triangle.
(China Beijing High School Grade 1 Mathematics Competition, 1991)
[Solution] Let the lengths of the legs be $a, b$, and the hypotenuse be $c$, where $a, b$, and $c$ are positive integers. From the problem,
$$
\begin{array}{l}
\left\{\begin{array}{l}
a^{2}+b^{2}=c^{2}, \\
c-b=1575, \\
ac-b=1575,105 k>1575, k>15.
\end{array}\right.
\end{array}
$$
Thus,
$$
k=16,17,18.
$$

Since $c-b=1575$ is odd, $c$ and $b$ have different parities, so $c+b$ is odd, and thus $a$ is odd.
Given that $a=105 k$ is odd, we have $\quad k=17$.
Therefore,
$$
a=105 \cdot 17=1785.
$$

Thus,
$$
\left\{\begin{array}{l}
c+b=2023, \\
c-b=1575.
\end{array}\right.
$$

Solving for the hypotenuse,
$$
c=1799.
$$