$10 \cdot 3$  Try to find the six-digit integer $\overline{a b c d e f}, \overline{a b c d e f} \cdot 3=$ $\overline{e f a b c d}$. Here $a, b, c, d, e, f$ represent different digits, and $a, e \neq 0$.
(Wuhan, Hubei Province, China Math Summer Camp, 1987)
[Solution] Let $\overline{a b c d}=x, \overline{e f}=y$. According to the problem,

that is
$$
(100 x+y) \cdot 3=10000 y+x,
$$

since
$$
23 x=769 y \text {. }
$$

we can assume
$$
(23,769)=1 \text {, }
$$

Since $y$ is a two-digit integer, $k$ can only take $1,2,3,4$.
When $k=1$, $x=769$ is not a four-digit number.
When $k=2$, $y=46, x=1538$, i.e., 153846;
When $k=3$, $y=69, x=2307$, i.e., 230769;
When $k=4$, $y=92, x=3076$, i.e., 307692.
The required six-digit numbers are 153846, 230769, and 307692.