11. In the following equation, the same letter represents the same digit, and different letters represent different digits, then
$\mathbf{A}=$ $\qquad$ ,$B=$ $\qquad$ , $\mathrm{C}=$ $\qquad$ , $\mathbf{D}=$ $\qquad$ $1^{\circ}$
【Answer】 $\mathrm{A}=9, \mathrm{~B}=8, \mathrm{C}=0, \mathrm{D}=1$
【Solution】It is easy to get $\mathrm{D}=1$, and $\mathrm{A}$ is at least 5 (otherwise, a carry would not occur). We can then discuss the cases.
1 ) If $\mathrm{A}=5$, from the units place, we get $\mathrm{B}=0$; from the tens place, we easily get $\mathrm{C}=4$; at this point, the hundreds place leads to a contradiction.
2 ) If $\mathrm{A}=6$, from the units place, we get $\mathrm{B}=2$; from the tens place, we easily get $\mathrm{C}=3$; at this point, the hundreds place leads to a contradiction.
3 ) If $\mathrm{A}=7$, from the units place, we get $\mathrm{B}=4$; from the tens place, we easily get $\mathrm{C}=2$; at this point, the hundreds place leads to a contradiction.
4 ) If $\mathrm{A}=8$, from the units place, we get $\mathrm{B}=6$; from the tens place, we easily get $\mathrm{C}=1$; at this point, the hundreds place leads to a contradiction.
5 ) If $\mathrm{A}=9$, from the units place, we get $\mathrm{B}=8$; from the tens place, we easily get $\mathrm{C}=0$; the hundreds place satisfies the condition.
Therefore, $\mathrm{A}=9, \mathrm{~B}=8, \mathrm{C}=0, \mathrm{D}=1$.