$74.1+2^{3}+3^{3}+\cdots+2017^{3}$ 的末位数字是

The last digit of $74.1+2^{3}+3^{3}+\cdots+2017^{3}$ is
Answer: 9
Solution: We are asked for the last digit, so we only need to consider the last digit of each addend.
Since the last digits of the cubes of $1,2,3,4,5,6,7,8,9,0$ are
$$
\begin{array}{c}
1,8,7,4,5,6,3,2,9,0, \\
2017 \div 10=201 \ldots \ldots 7, \\
1+8+7+4+5+6+3+2+9+0=45, \\
45 \times 201+(1+8+7+4+5+6+3)=9079,
\end{array}
$$

Therefore, the last digit of $1+2^{3}+3^{3}+\cdots+2017^{3}$ is 9.