4. The last two digits of the integer $\left[\frac{10^{93}}{10^{31}+3}\right]$ are $\qquad$ (write the tens digit first, then the units digit; where $[x]$ denotes the greatest integer not exceeding $x$).
4.
$$
\begin{aligned}
{\left[\frac{10^{93}}{10^{31}+3}\right] } & =\left[\frac{\left(10^{31}\right)^{3}+3^{3}-3^{3}}{10^{31}+3}\right] \\
& =\left(10^{31}\right)^{2}-3 \times 10^{31}+3^{2}-1 \\
& =\left(10^{31}\right)\left(10^{31}-3\right)+8
\end{aligned}
$$

So its last two digits are 08.