11. Find the last three digits of $2016^{3}+2017^{3}+2018^{3}+\ldots+3014^{3}$.
Answer: 625
Solution: Note that $\{2016,2017,2018, \ldots, 3014,3015\}$ comprise 1000 consecutive integers, and thus produce all possible residues modulo 1000. Thus,
$$
\begin{aligned}
2016^{3}+2017^{3}+2018^{3}+\ldots+3014^{3}+3015^{3} & \equiv 0^{3}+1^{3}+2^{3}+\ldots+999^{3} \quad(\bmod 1000) \\
& =\left(\frac{999 \cdot 1000}{2}\right)^{2} \\
& =999^{2} \cdot 500^{2} \\
& =999^{2} \cdot 250000 \\
& \equiv 0(\bmod 1000)
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
2016^{3}+2017^{3}+2018^{3}+\ldots+3014^{3} & =2016^{3}+2017^{3}+2018^{3}+\ldots+3014^{3}+3015^{3}-3015^{3} \\
& \equiv 0-3015^{3}(\bmod 1000) \\
& \equiv 1000-15^{3} \quad(\bmod 1000) \\
& \equiv 1000-375 \quad(\bmod 1000) \\
& \equiv 625 \quad(\bmod 1000)
\end{aligned}
$$