10. The last two digits of the integer $\left[\frac{10^{93}}{10^{31}+3}\right]$ are $\qquad$ . (Write the tens digit first, followed by the units digit, where $[x]$ denotes the greatest integer less than or equal to $x$)
$$
\left[\frac{10^{93}}{10^{31}+3}\right]=\left[\frac{10^{93}+27}{10^{31}+3}-\frac{27}{10^{31}+3}\right]=\left[10^{62}-3 \cdot 10^{31}+9-\frac{27}{10^{31}+3}\right]
$$
$=10^{62}-3 \cdot 10^{31}+8$, so the last two digits are 08.