$\triangle$ Let's look at the last column: it contains the same digit A. What is it equal to? Only zero.

Now let's pay attention to the second column: there is a similar situation with the digit O. Hence, O is equal to zero or 9. But the first possibility is ruled out; thus, $\mathrm{O}=9$.

To find K, consider the first column. Clearly, K is not zero and does not exceed 4. Then $\mathrm{K}$ takes one of the values $1,2,3,4$. Let's consider four cases.

1) Let $\mathrm{K}=1$.

We get that in the third column, $Л=9$, since in the second column it should be $9+9+1=19$. But then Д $=0$, which is impossible.

2) Let $\mathrm{K}=2$.

Substitute into the puzzle the values $\mathrm{K}=2, \mathrm{~A}=0, \mathrm{O}=9$.

$$
\begin{aligned}
& 2920 \\
& +\begin{array}{c}
2920 \\
29 Л 0
\end{array} \\
& \hline \text { В9Д0. }
\end{aligned}
$$

From the third column

$$
2+\text { Л = } 10+\text { Д, } \quad \text { Л = } 8+\text { Д. }
$$

Hence, $Д=0$ or $Д=1$, respectively, $Л=8$ or $Л=9$. But both these possibilities are excluded.

3) Let $\mathrm{K}=3$.

We get:

$$
\begin{array}{r}
3930 \\
+\begin{array}{l}
39 Л 0 \\
\text { В9Д0. }
\end{array} \\
\hline
\end{array}
$$

Then

$$
3+Л=10+Д, \quad Л=7+Д,
$$

which means, $Д=1, Л=8$. In addition, $В=7$.

4) Let $К=4$.

Therefore, $\mathrm{B}=9$. But the latter is impossible.

Thus, the solution is obtained only in the third case.

Answer: $3930+3980=7910$.
53. Solve the rebus:

$$
\begin{aligned}
& \text { COCA } \\
&+ \\
& \text { COLA } \\
& \hline \text { WATER. }
\end{aligned}
$$