5. $[x]$ represents the greatest integer not exceeding the real number $x$, for example $[3]=3$, $[2.7]=2,[-2.2]=-3$, then the last two digits of $\left[\frac{10^{93}}{10^{31}+3}\right]$ are . $\qquad$
5. 08 Detailed Explanation: Since $\frac{10^{93}}{10^{31}+3}=\frac{\left(10^{31}\right)^{3}+3^{3}}{10^{31}+3}-\frac{3^{3}}{10^{31}+3}=10^{62}-3 \times 10^{31}+9-\frac{3^{3}}{10^{31}+3}$, therefore $10^{62}-3 \times 10^{31}+8<\frac{10^{93}}{10^{31}+3}<10^{62}-3 \times 10^{31}+9$, hence $\left[\frac{10^{93}}{10^{31}+3}\right]=10^{62}-3 \times 10^{31}+8$, which means the last two digits are 08.