3. Replace the letters with digits so that the equality is correct. Different letters correspond to different digits.

$$
\begin{array}{ccccc} 
& \text { A } & \text { B } & \text { A } & \text { C } \\
& + & \text { B } & \text { A } & \text { C } \\
\text { K } & \text { C } & \text { K } & \text { D } & \text { C }
\end{array}
$$
3. Since we are adding a four-digit number to a three-digit number and getting a five-digit number, the first digit in the sum is one $(K=1)$.

$$
\begin{array}{rrrr} 
& \text { B } & \text { C } \\
+ & B & A & C \\
1 & C & 1 & \text { C }
\end{array}
$$

When adding the numbers in the units place, the result should be the same number or the sum of this number and ten. This can only happen in one case, if $\mathrm{C}=0$.

$$
\begin{array}{rrrr}
\text { A } & \text { B } & \text { A } \\
+ & \text { B } & \text { A } 0 \\
1 & 0 & 1 & \text { D }
\end{array}
$$

Since $10=\mathrm{A}+1$, we can determine $\mathrm{A}=10-1=9$. We get the following sum:

$$
\begin{array}{rrrr}
9 & \text { B } & 9 & 0 \\
+ & \text { B } & 9 & 0 \\
1 & 0 & 1 & \text { D }
\end{array}
$$

Now we can determine what D is $(9+9=18, D=8)$.

$$
\begin{aligned}
& 9 \text { B } 9 \\
& + \text { B } 90 \\
& \begin{array}{lllll}
1 & 0 & 1 & 8 & 0
\end{array}
\end{aligned}
$$

Finally, we find $\mathrm{B}(\mathrm{B}+\mathrm{B}=11-1,2 \mathrm{~B}=10, \mathrm{~B}=5)$. We get the following sum:

$$
\begin{aligned}
& \begin{array}{llll}
9 & 5 & 9 & 0
\end{array} \\
& +590 \\
& \begin{array}{lllll}
1 & 0 & 1 & 8 & 0
\end{array}
\end{aligned}
$$

Answer: $\mathrm{A}=9, \mathrm{~B}=5, \mathrm{C}=0, \mathrm{~D}=8, \mathrm{~K}=1$.