$14 \cdot 35$ Find the units digit of $\left[\frac{10^{20000}}{10^{100}+3}\right]$.
(47th Putnam Mathematical Competition, 1986)
[Solution] Let $10^{100}=t$, then
$$
\begin{aligned}
A & =\left[\frac{10^{20000}}{10^{100}+3}\right] \\
& =\left[\frac{t^{200}}{t+3}\right] \\
& =\left[\frac{t^{200}-3^{200}}{t+3}+\frac{3^{200}}{t+3}\right],
\end{aligned}
$$

Since $3^{200}=9^{100}<t+3$, we have
$$
0<\frac{3^{200}}{t+3}<1 \text {. }
$$

Also, because $t+3 \nmid t^{200}-3^{200}$, then
$$
\begin{aligned}
A & =\frac{t^{200}-3^{200}}{t+3} \\
& =\frac{t^{200}-81^{50}}{t+3} .
\end{aligned}
$$

Since the last digit of the denominator $t+3$ is 3, and the last digit of the numerator $t^{200}-81^{50}$ is 9,
the last digit of $A$ is 3.