5. The last two digits of the integer $\left[\frac{10^{1995}}{10^{95}+3}\right]$ are ( ).
(A) 10
(B) 01
(C) 00
(D) 20
5. (C).
$$
\begin{array}{l}
\because A=\frac{10^{1995}}{10^{95}+3}=\left(10^{95 \times 20}-3 \times 10^{95 \times 19}+\cdots\right. \\
\left.-3^{19} \times 10^{95}+3^{20}\right)-\frac{3^{21}}{10^{95}+3}, \\
\begin{aligned}
\therefore[A]= & 10^{1900}-3 \times 10^{1805}+\cdots-3^{19} \times 10^{95} \\
& \quad+3^{20}-1 .
\end{aligned}
\end{array}
$$

Since $3^{20}-1=(10-1)^{10}-1 \equiv 1-1 \equiv 0(\bmod 100)$, it follows that the last two digits of $[A]$ are 00.