10.4 For each integer from 10 to 2021, we found the product of the digits, and then added all the obtained results. What is the sum that was obtained?
Solution: Consider the product $(1+2+3+\cdots+9) \cdot(0+1+2+\cdots+9)$. If we expand the brackets, we get the products of pairs of digits that form all two-digit numbers. Similarly, in the product

$$
(1+2+3+\cdots+9) \cdot(0+1+2+\cdots+9) \cdot(0+1+2+\cdots+9)
$$

after expanding the brackets, we get all combinations of digits from three-digit numbers, and

$$
1 \cdot(0+1+2+\cdots+9) \cdot(0+1+2+\cdots+9) \cdot(0+1+2+\cdots+9)
$$

gives all combinations for numbers from 1000 to 1999. Numbers from 2000 to 2021 give a zero product of digits. Therefore, the answer is $45^{2}+45^{3}+45^{3}=184275$.

Criteria:

- The answer is given by the formula $45^{2}+45^{3}+45^{3}$ and the final answer in decimal form is not calculated - no points deducted
- Only the sum of the products for two-digit numbers is correctly calculated - 1 point