
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:

$$(2 \cdot 5) \cdot(10) \cdot(12 \cdot 15) \cdot(25 \cdot 4) \cdot(20 \cdot 30) \cdot(35 \cdot 14) \ldots \Rightarrow$$
since each of these grouped products ends with 0, the number will end with 8 zeros $\Rightarrow B=0$.

Rewrite the expression into the following products:

$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 = \ldots$$ 
product ends with 8 (without zeros)

$$11 \cdot 12 \cdot 13 \cdot 14 \cdot 15 \cdot 16 \cdot 17 \cdot 18 \cdot 19 = \dots$$ 
product ends with 4 (without zeros)

$$21 \cdot 22 \cdot 23 \cdot 24 \cdot 25 \cdot 26 \cdot 27 \cdot 28 \cdot 29 = \ldots$$ 
product ends with 4 (without zeros)

$$31 \cdot 32 \cdot 33 \cdot 34 \cdot 35 = \ldots$$ 
product ends with 4 (without zeros)

$$8 \cdot 4 \cdot 4 \cdot 4 = \ldots$$ 
product ends with 2 $\Rightarrow \mathrm{A}=2$

Consider the characteristics of divisibility by 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:

$$71+C-68-D \Rightarrow 11 \Rightarrow 3+C-D \Rightarrow 11$$

Hence, $C-D=-3$ or $C-D=8$.

By the divisibility rule for 9:

$$68+D+C+71 \vdots 9 \Rightarrow 139+D+C \vdots 9$$

Hence, $D+C=5$ or $D+C=14$.

Set up the system of equations:
$$\left\{\begin{array}{c}
C-D=-3 \\
C+D=5
\end{array} \Rightarrow 2C=2 \Rightarrow C=1 \Rightarrow D=4\right.$$

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