In the lottery, the following numbers were drawn in a week: $\overline{A B}, \overline{B C}, \overline{C A}, \overline{C B}, \overline{C D}$ (A, B, C, D are digits, the numbers are arranged in ascending order). The sum of the five numbers is $\overline{B C C}$, the product of the third number and the second number is $\overline{B B E C}$, and the product of the third number and the fifth number is $\overline{E C C D}$. Determine the digits.
I. solution. Due to the announced increasing order, $1 \leq A0$ implies $D=5, \overline{C D}$ is odd, hence $\overline{C A}-1$ is divisible by 20. Then $A=1$, and $C$ is even, so $\overline{E C}$ is also even, both sides are divisible by 8, therefore $\overline{C A}-1$ is divisible by 40, so $C$ can only be 4 and 8. In the case of $C=4$, the product $40 \cdot 45=100 \cdot 18$ is not in the form (1), whereas for $C=8$, $80 \cdot 85=100 \cdot 68$ is in the form and $E=6$.

Now, from the first product requirement, $\overline{C A} \cdot \overline{B C}=\overline{B B E C}, 81(10 B+8)=1100 B+68$. From this, $B=2$, and the extracted numbers: $12,28,81,82,85$, also meet the sum requirement.

Remark. The last requirement does not include $B$, so we had to use an additional requirement. Similarly, in the III. solution, where we start from the first product, we calculate $D$ from the second product.

III. solution. Transform the first product requirement as follows:

$$
\begin{gathered}
\overline{C A} \cdot \overline{B C}=\overline{B B E C}, \quad 10 B \cdot \overline{C A}+C \cdot \overline{C A}=1100 B+\overline{E C} \\
B=\frac{C \cdot \overline{C A}-\overline{E C}}{10(110-\overline{C A})} \leqq \frac{8 \cdot 86-\overline{E C}}{10(110-86)}A \text{ (we did not consider the } A \text{ requirement in the first transformation.)} \text{ According to this, } A=1 \text{ and } B=2.
\end{gathered}
$$

On the other hand, $E>B$, and thus $E \geq 3$, because the fifth number is larger than the second, so the second product is larger than the first product, and the first digits of the products are different.

Based on the above, (2) can be transformed as follows:

$$
\begin{gathered}
(10 C+1)(20+C)=2200+10 E+C \geqq 2230+C \\
C^{2}+20 C-221 \geq 0
\end{gathered}
$$

from which $C \geq 8$ or $C \leq-28$. Only $C=8$ fits, because $C<9$, so the value of the first product is $81 \cdot 28=2268$, thus $E=6$.

Accordingly, the second product falls between 6880 and 6890, dividing these by the third number, 81, we get a lower and an upper limit for the fifth number: 84.9 (rounded down) and 85.1 (rounded up), so the fifth number is 85, i.e., $D=5$.