8. In a 1000-meter race, an hourglass is leaking sand at a constant rate, and the sand that leaks out falls into a cup (this hourglass starts leaking after the race has been going on for some time). Xiao Ming runs at a constant speed. When he reaches 200 meters, the $\mathrm{a}$-th grain of sand falls into the cup. When he reaches 300 meters, the $\overline{b c}$-th grain of sand falls into the cup. When he reaches 400 meters, the $\overline{d e}$-th grain of sand falls into the cup. When he reaches 500 meters, the $\overline{f g}$-th grain of sand falls into the cup $(a, b, c, d, e, f, g$ are digits from $0$ to $9$, and their values can be the same). We find that: (1) $\mathrm{a}$ is a multiple of 2, (2) $\overline{b c}$ is a prime number; (3) $\overline{d e}$ is a multiple of 5; (4) $\overline{f g}$ is a multiple of 3. What is the four-digit number $\overline{d e b c}=$ $\qquad$ (if there are multiple solutions, write all of them on the line).
【Analysis】From the fact that the hourglass leaks sand at a uniform rate, we know $\overline{f g}-\overline{d e}=\overline{d e}-\overline{b c}=\overline{b c}-\mathrm{a}$
Therefore, let's assume $\overline{b c}=\mathrm{a}+\mathrm{k}, \overline{d e}=\mathrm{a}+2 \mathrm{k}, \overline{f g}=\mathrm{a}+3 \mathrm{k}$,
From $3|\overline{f g} \rightarrow 3| \mathrm{a}+3 \mathrm{k} \rightarrow 3 \mid \mathrm{a}$, and since $\mathrm{a}$ is a multiple of 2, $\mathrm{a}$ must be a common multiple of 2 and 3, so $\mathrm{a}=0$ or 6

If $\mathrm{a}=0$, then by $5|\overline{d e} \rightarrow 5| \mathrm{a}+2 \mathrm{k} \rightarrow 5 \mid \mathrm{k}$, i.e., $5 \mid \overline{b c}$, which contradicts the fact that $\overline{b c}$ is a prime number. Therefore, $\mathrm{a}=6$

From $\overline{b c}=6+\mathrm{k} \rightarrow \mathrm{k} \geqslant 4$, and from $\overline{f g}=6+3 \mathrm{k} \rightarrow \mathrm{k} \leqslant 31$
From $5|\overline{d e} \rightarrow 5| 6+2 \mathrm{k} \rightarrow \mathrm{k}$ must end in 2 or 7
Since $\overline{c c}=6+\mathrm{k}$ is a prime number, $\mathrm{k}$ must be odd and not a multiple of 3
Thus, $\mathrm{k}$ must end in 7, and be between 4 and 31, and not a multiple of 3
Therefore, the possible values for $\mathrm{k}$ are 7, 17
When $\mathrm{k}=7$, $\mathrm{a}=6, \overline{b c}=13, \overline{d e}=20, \overline{f g}=27$, which meets the requirements, so $\overline{d e b c}=2013$
When $\mathrm{k}=17$, $\mathrm{a}=6, \overline{b c}=23, \overline{d e}=40, \overline{f g}=57$, which meets the requirements, so $\overline{d e b c}=4023$
In summary, $\overline{d e b c}=2013$ or 4023