3 . Let $n$ be a positive integer, $d$ be a digit among the 10 decimal digits, and suppose $\frac{n}{810}=0 . d 25 d 25 d 25 \cdots$, find $n$.


$$
\begin{array}{c}
3 . \because \frac{n}{810}=0 . d 25 d 25 d 25 \cdots, \\
\therefore 1000 \cdot \frac{n}{810}=d 25 . d 25 d 25 \cdots \\
\therefore \frac{999 n}{810}=1000 \cdot \frac{n}{810}-\frac{n}{810} \\
=d 25=100 d+25, \\
\therefore \quad 999 n=810(100 d+25),
\end{array}
$$

And $(750,37)=1$, so $37 \mid (4d+1)$, and since $d$ is a single digit, hence $d=9$, then $n=750$.