11. In the multiplication equation in the figure, different Chinese characters represent different digits, and the same Chinese characters represent the same digit. Fill in the appropriate digits in the squares of the equation to make the equation true. Then, the three-digit number represented by "中环杯" is $\qquad$

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
【Answer】 124

【Solution】Since $\left\{\begin{array}{l}\overline{a b c} \times 6 \text { is a three-digit number } \\ \overline{a b c} \times d, \overline{a b c} \times e \text { are both four-digit numbers }\end{array}\right.$, so $d, e$ are digits in the range $7 \sim 9$. From “$\overline{a b c} \times 6$ is a three-digit number”, we can easily deduce that $a=1$. Since “中环杯中环杯 $=$ 中环杯 $\times 1001$”, it follows that $1001 \mid \overline{a b c} \times \overline{d 6 e}$. Considering that $\overline{a b c}$ is only a three-digit number, it is impossible to satisfy $\left\{\begin{array}{l}7 \mid \overline{a b c} \\ 11 \mid \overline{a b c} \\ 13 \mid \overline{a b c}\end{array}\right.$, so at least one of $7,11,13$ must divide $\overline{d 6 e}$. We proceed with the following cases:
(1) If $7 \mid \overline{d 6 e}$, and given that $d, e$ are digits in the range $7 \sim 9$, we can easily deduce that only 868 satisfies the condition. In this case, 11 and 13 do not divide 868, so 11 and 13 must divide $\overline{a b c}=\overline{1 b c}$, so $\overline{1 b c}=11 \times 13=143$, and $143 \times 868=124124$, which meets our requirements.
(2) If $11 \mid \overline{d 6 e}$, and given that $d, e$ are digits in the range $7 \sim 9$, so $d+e-6=11 \Rightarrow d+e=17$, we can easily deduce that only 869 or 968 satisfy the requirement. Considering that 7 and 13 do not divide these two numbers, 7 and 13 must divide $\overline{a b c}=\overline{1 b c}$, so $\overline{1 b c}=7 \times 13 \times 2=182$, but $182 \times 6$ is not a three-digit number, which is a contradiction.
(3) If $13 \mid \overline{d 6 e}$, and given that $d, e$ are digits in the range $7 \sim 9$, we can easily deduce that only 767 satisfies the condition. Considering that 7 and 11 do not divide these two numbers, 7 and 11 must divide $\overline{a b c}=\overline{1 b c}$, so $\overline{1 b c}=7 \times 11 \times 2=154$. In this case, $154 \times 767=118118$, but the problem states that the three digits of “中环杯” are all different, so this does not meet the requirement.

In summary, there is only one final answer, which is “中环杯 $=124$”.