
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
In the expression, the following products are contained:

\[
(2 \cdot 5) \cdot (10) \cdot (12 \cdot 15) \cdot (25 \cdot 4) \cdot (20 \cdot 30) \cdot (35 \cdot 14) \ldots
\]

Since each of these grouped products ends in 0, the number will end with 8 zeros, implying \( B = 0 \).

Let's break the expression into the following products:

\[
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 = \ldots 8 \ \text{(ends in 8 without zeros)}
\]

\[
11 \cdot 12 \cdot 13 \cdot 14 \cdot 15 \cdot 16 \cdot 17 \cdot 18 \cdot 19 = \ldots 4 \ \text{(ends in 4 without zeros)}
\]

\[
21 \cdot 22 \cdot 23 \cdot 24 \cdot 25 \cdot 26 \cdot 27 \cdot 28 \cdot 29 = \ldots 4 \ \text{(ends in 4 without zeros)}
\]

\[
31 \cdot 32 \cdot 33 \cdot 34 \cdot 35 = \ldots \ \text{(ends in 4 without zeros)}
\]

\[
8 \cdot 4 \cdot 4 \cdot 4 = \ldots 2 \ \text{(ends in 2)} \ \Rightarrow \ \text{A} = 2
\]

Consider the divisibility rules for 11 and 9:

Sum of digits at even positions:

\[
3 + 1 + 7 + 6 + 3 + 6 + D + 9 + 9 + 6 + 6 + 1 + 3 + 5 + 3 = 68 + D
\]

Sum of digits at odd positions:

\[
1 + 3 + 3 + 4 + 9 + 6 + 8 + C + 4 + 2 + 6 + 6 + 5 + 3 + 7 + 2 + 2 = 71 + C
\]

By the divisibility rule for 11:

\[
71 + C - (68 + D) = 11, \ 3 + C - D \vdots 11
\]

Hence, \( C - D = -3 \) or \( C - D = 8 \).

By the divisibility rule for 9:

\[
68 + D + C + 71 \equiv 9, \ 139 + D + C \equiv 9
\]

Hence, \( D + C = 5 \) or \( D + C = 14 \).

Construct the system of equations:
\[
\begin{cases}
C - D = -3 \\
C + D = 5
\end{cases}
\]
Solving this gives:
\[
2C = 2 \ \Rightarrow \ C = 1 \ \Rightarrow D = 4
\]

Present the answer in LaTex format: \boxed{Your answer}
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