11. Given a four-digit number $\overline{A B C D}$ that satisfies: $\overline{A B C D}+\overline{A B} \times \overline{C D}$ is a multiple of 1111, then the minimum value of $\overline{A B C D}$ is . $\qquad$
【Answer】 1729
【Solution】 $\overline{A B C D}+\overline{A B} \times \overline{C D}=100 \overline{A B}+\overline{C D}+\overline{A B} \times \overline{C D}=(\overline{A B}+1) \times(\overline{C D}+100)-100$, thus leading to $(\overline{A B}+1) \times(\overline{C D}+100) \equiv 100(\bmod 1111)$, so $(\overline{A B}+1) \times(\overline{C D}+100)=1211, 2322, 3433$,
(1) When $(\overline{A B}+1) \times(\overline{C D}+100)=1211$, at this time $\overline{A B}+1=\frac{1211}{\overline{C D}+100} \leq \frac{1211}{100}=12.11$, so $\overline{A B} \leq 11$, so $\overline{A B}=11$ or 10, but these two numbers are clearly not factors of 1211;
(2) When $(\overline{A B}+1) \times(\overline{C D}+100)=2322$, considering $2322=2 \times 3^{3} \times 43$, so $2322=18 \times 129$, at this time $\left\{\begin{array}{l}\overline{A B}+1=18 \\ \overline{C D}+100=129\end{array} \Rightarrow \overline{A B C D}=1729\right.$.

Next, we need to prove that 1729 is already the minimum value. Assume it is not the minimum value, and there exists $\overline{A B C D}<1729$, satisfying $(\overline{A B}+1) \times(\overline{C D}+100)=1111 k+100$, at this time $\overline{A B}=10 \sim 16$, so $(\overline{A B}+1) \times(\overline{C D}+100) \leq(16+1) \times(99+100)=3383$. When $k \geq 3$, at this time $1111 k+100$ is already greater than 3383, so $k \leq 2$. And for the case of $k \leq 2$, we have already discussed it, so there does not exist $\overline{A B C D}<1729$.

In summary, the minimum value required for this problem is 1729.