10.34 Let $n$ be a positive integer, and $d$ be one of the digits in the decimal system. If $\frac{n}{810} = \overline{0.d25d25d25\cdots}$, find $n$.
(7th American Invitational Mathematics Examination, 1989)
[Solution] Since
$$
\frac{n}{810}=\overline{0 . d 25 d 25 d 25 \cdots},
$$

then
$$
1000 \cdot \frac{n}{810}=\overline{d 25 . d 25 d 25 \cdots} \text {, }
$$

thus
$$
1000 \cdot \frac{n}{810}-\frac{n}{810}=\overline{d 25} \text {, }
$$
$\frac{999 n}{810}=\overline{d 25}$,
$\frac{999 n}{810}=100 d+25$,
$999 n=810(100 d+25)$,
$37 n=750(4 d+1)$.
Since $(37,750)=1$, then
$$
37 \mid 4 d+1 \text {. }
$$

Also, because $d \in\{0,1,2, \cdots, 9\}$, then $d=9$,
thus $n=750$.