Example 1 (1993 National High School League Question) The last two digits of $\left[\frac{10^{93}}{10^{31}+3}\right]$ are $\qquad$ (where [x] denotes the greatest integer not greater than $x$.)
Solution: Fill in 08. Reason: Since $10^{93}=\left(10^{31}\right)^{3}+3^{3}-3^{3}=\left(10^{3}+3\right)\left[\left(10^{31}\right)^{2}-10^{31} \cdot 3+\right.$ $\left.3^{2}\right]-3^{3}$, therefore

the original expression $=\left(10^{31}\right)^{2}-10^{31} \cdot 3+3^{2}-1=10^{31}\left[10^{31}-3\right]+8$. Hence, the last two digits of the number are 08.