G4.2 Given that the unit digit of $1^{3}+2^{3}+3^{3}+\ldots+2001^{3}$ is $b$, find the value of $b$.
Arrange the numbers in groups of 10 in ascending order, the unit digit of sum each group is the same (except the last number, $2001^{3}$ ).
$$
\begin{aligned}
1^{3}+2^{3}+\cdots+10^{3} & \equiv++8+7+4+5+6+3+2+9+0(\bmod 10) \\
& \equiv 5(\bmod 10) \\
1^{3}+2^{3}+\cdots+2000^{3} & +2001^{3} \equiv 200(5)+1(\bmod 10)
\end{aligned}
$$

So $b=1$