1. (1993 National High School Mathematics Competition) What are the last two digits of the integer $\left[\frac{10^{93}}{10^{31}+3}\right]$? (Write the tens digit first, then the units digit)
1. Let $10^{31}=t$, then
$$
\begin{aligned}
{\left[\frac{10^{93}}{10^{31}+3}\right] } & =\left[\frac{t^{3}}{t+3}\right]=\left[\frac{t^{3}+3^{3}}{t+3}-\frac{3^{3}}{t+3}\right]=t^{2}-3 t+3^{2}+\left[-\frac{3^{3}}{t+3}\right]=t^{2}-3 t+3^{2}-1 \\
& =t(t-3)+8=10^{31}\left(10^{31}-3\right)+8 .
\end{aligned}
$$

Therefore, the last two digits of $\left[\frac{10^{93}}{10^{31}+3}\right]$ are 08.