10. (5 points) The sum of three numbers, A, B, and C, is 2017. A is 3 less than twice B, and B is 20 more than three times C. Then A is $\qquad$ .
【Analysis】If乙is 20 more than 3 times of丙, then the number乙can be expressed as丙$\times 3+20$. If甲is 3 less than 2 times of乙, then the number甲is 2 times 3 times of丙plus $20 \times 3$. Therefore, the sum of the three numbers is 1 plus 2 times 3 plus 3 times of丙, plus $(20 \times 3-3)$. By subtracting $(20 \times 3-3)$ from the sum of the three numbers, we get 1 plus 2 times 3 plus 3 times of丙, which allows us to find the value of丙, and subsequently the values of乙and甲.

【Solution】Solution: For丙: $(2017-20 \times 3+3) \div(1+2 \times 3+3)$
$$
\begin{array}{l}
=(2017-57) \div 10 \\
=1960 \div 10 \\
=196,
\end{array}
$$

For乙: $196 \times 3+20=608$,
For甲: $608 \times 2-3=1213$,
Answer: 甲is 1213.
Therefore, the answer is: 1213.