In the addition below, the same letters represent the same digit, and different letters represent different digits. Find the number $A B C D E$.

$A B C D E$ $B C D E$ $C D E$ $D E$ $A A A A A$
Note that $5 \times E$ is a multiple of 5 and, in this case, ends in $A$. Since $A$ cannot be 0, it follows that $A=5$ and $E$ is odd. Observe that $E$ cannot be 1, because in that case, $4 D=5$, which is impossible for digits. Therefore, $E=3, 5, 7$ or 9. Let's analyze each of these possibilities.

If $E=3$, then $4 D+1$ ends in 5, and therefore, $D=1$ or $D=6$;

If $E=5$, then $4 D+2$ is even and ends in 5, which is impossible;

If $E=7$, then $4 D+3$ ends in 5, and therefore, $D=3$ or $D=8$;

If $E=9$, then $4 D+4$ is even and ends in 5, which is impossible.

Thus, we are left with the following four cases, according to $(D, E)$ being given by $(1,3)$, $(6,3)$, $(3,7)$, or $(8,7)$.

| $5 B C 13$ | $5 B C 63$ | $5 B C 37$ | $5 B C 87$ |
| ---: | ---: | ---: | ---: |
| $B C 13$ | $B C 63$ | $B C 37$ | $B C 87$ |
| $C 13$ | $C 63$ | $C 37$ | $C 87$ |
| 13 | 63 | 37 | 87 |
| 3 | 3 | 7 |  |
|  |  | 55555 | 55555 |

1st Case: If $D=1$ and $E=3$, then $3 C$ ends in 5, and since $C$ denotes a digit, the only option is $C=5=A$, which cannot occur.

2nd Case: If $D=6$ and $E=3$, then $3 C+2$ ends in 5, so $3 C$ ends in 3. Since $C$ denotes a digit, the only option is $C=1$, resulting in $2 B=5$, which is impossible for a digit.

3rd Case: If $D=3$ and $E=7$, then $3 C+1$ ends in 5, so $3 C$ ends in 4. Since $C$ denotes a digit, the only option is $C=8$, resulting in $2 B+2=5$, which is impossible.

4th Case: If $D=8$ and $E=7$, then $3 C+3$ ends in 5, so $3 C$ ends in 2. Since $C$ denotes a digit, the only option is $C=4$, resulting in $2 B+1=5$, with which $B=2$.

Thus, the only solution is $A B C D E=52487$.