13. (12 points) A three-digit number has different digits in each place. Writing 2 at the beginning of this three-digit number forms a four-digit number; writing 2 at the end of this three-digit number also forms a four-digit number; the difference between these two four-digit numbers is 945. What is this three-digit number?
A. 117
B. 327
C. 219
D. 312
【Solution】Solution: Let the three-digit number be $\overline{\mathrm{abc}}$. According to the problem, we have $\overline{2 \mathrm{abc}}-\overline{\mathrm{abc2}}=945$ or $\overline{\mathrm{abc2}}-\overline{2 \mathrm{abc}}=945$. By treating $a b c$ as a group, we get:
$$
\begin{array}{l}
\text { When: } \overline{2 a b c}-\overline{a b c 2}=945, \\
2000+\overline{a b c}-10 \overline{a b c}-2=945, \\
1998-9 \overline{a b c}=945,
\end{array}
$$
$\overline{a b c}=117$ (contradicts the condition that the digits are distinct),
$$
\begin{array}{l}
\text { When: } \overline{a b c 2}-\overline{2 a b c}=945, \\
10 \overline{a b c}+2-2000-\overline{a b c}=945, \\
9 \overline{a b c}-1998=945, \\
\overline{a b c}=327 \text { (satisfies the condition). }
\end{array}
$$

Therefore, the answer is: $B$.