6. The expression contains the following products:

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

Let's break down the expression into the following products:

$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9=\ldots 8$ the product ends in 8 (no zeros)

$11 \cdot 12 \cdot 13 \cdot 14 \cdot 15 \cdot 16 \cdot 17 \cdot 18 \cdot 19=\ldots$ the product ends in 4 (no zeros)

$21 \cdot 22 \cdot 23 \cdot 24 \cdot 25 \cdot 26 \cdot 27 \cdot 28 \cdot 29=\ldots 4$ the product ends in 4 (no zeros)

$31 \cdot 32 \cdot 33 \cdot 34 \cdot 35=\ldots 4$ the product ends in 4 (no zeros)

$8 \cdot 4 \cdot 4 \cdot 4=\ldots .2$ the product ends in $2 \Rightarrow A=2$

Consider the divisibility rules for 11 and 9:

Sum of digits in even positions:

$3+1+7+6+3+6+D+9+9+6+6+1+3+5+3=68+D$

Sum of digits in 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:

$71+C-68-D: 11,3+C-D: 11$

From this, $C-D=-3$ or $C-D=8$.

By the divisibility rule for 9:

$68+D+C+71 \vdots 9,139+D+C: 9$

From this, $D+C=5$ or $D+C=14$.

Let's form a system of equations: $\left\{\begin{array}{c}C-D=-3 \\ C+D=5\end{array} \Rightarrow 2C=2 \Rightarrow C=1 \Rightarrow D=4\right.$

Note: if we consider the other cases, then C is either fractional or greater than 10, which means this solution is the only one.
Answer: $A=\mathbf{2}, \mathrm{B}=\mathbf{0}, \mathrm{C}=\mathbf{1}, \mathrm{D}=\mathbf{4}$.

| 20 points | Correct answer obtained with justification |
| :---: | :--- |
| 15 points | Three out of four digits are correctly found. |
| 10 points | Two out of four digits are correctly found. |
| 5 points | One out of four digits is correctly found. |
| 0 points | The solution does not meet any of the criteria listed above |