11. (12 points) Junjun is looking at an incorrect one-digit multiplication equation, $A \times B=\overline{\mathrm{CD}}$ (where $A, B, C, D$ represent different digits), and the clever Junjun discovers that if only one digit is changed, there are 3 ways to correct it; if only the order of $A, B, C, D$ is changed, it can also be corrected. What is $A+B+C+D=$ $\qquad$ .
11. (12 points) Junjun is looking at an incorrect one-digit multiplication equation, $A \times B=\overline{\mathrm{CD}}$ (where $A, B, C, D$ represent different digits), and the clever Junjun discovers that if only one digit is changed, there are 3 ways to correct it; if only the order of $A, B, C, D$ is changed, it can also be corrected. What is $A+B+C+D=$ $\qquad$.

【Solution】Solution: According to the analysis, the only situations that can be transformed into two pairs of numbers whose product is equal are:
(1) $3 \times 4=2 \times 6=12$;
(2) $4 \times 4=2 \times 8=16$;
(3) $3 \times 6=2 \times 9=18$;
(4) $4 \times 6=3 \times 8=24$;
(5) $4 \times 9=6 \times 6=36$,
Clearly, (2) and (5) do not meet the requirements. Changing the order to correct it means that these digits remain unchanged, only their positions change.

In the first group (1), if $A=3, B=6$ or $A=3, B=2$, then $\overline{C D}=12$. Changing one digit results in: $3 \times 4=12$ or $2 \times 6=12$ or $3 \times 6=18$,
However, changing the order, the equation $3 \times 6=12$ cannot be corrected no matter how the order is changed, so (1) is excluded. In the third group (3), if $A=6, B=2$ (or $A=2, B=6$), then $\overline{\mathrm{CD}}=18$,
$6 \times 2=18$ can be changed to: $6 \times 3=18$; $9 \times 2=18$; $6 \times 2=12$, changing the order results in: $2 \times 8=16$, which meets the requirements. If $A=9, B=3$ (or $A=3, B=9$), then $\overline{C D}=18$,
$9 \times 3=18$ can be changed to: $6 \times 3=18$; $9 \times 2=18$; (only two changes possible), so it is excluded. If $A=9, B=6, \overline{C D}=18$,
$9 \times 6=18$ can be changed to: $3 \times 6=18$; $9 \times 2=18$; (only two changes possible), so it is excluded.
In the fourth group (4), if $A=6, B=3, \overline{\mathrm{CD}}=24$, it can be changed to: $6 \times 4=24$; $3 \times 8=24$; (only two changes possible), so it is excluded.

If $A=6, B=8, 6 \times 8=24$ can be changed to: $6 \times 4=24$; $3 \times 8=24$; (only two changes possible), so it is excluded. Therefore: $A=6, B=2$ (or $A=2, B=6$), $\overline{\mathrm{CD}}=18$, $A+B+C+D=6+2+1+8=17$.