Rightmost digit of $ \left \lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor $ How could I find $$ 0 \leq a \leq 9 $$ such that
$$ \left \lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor \equiv a \mod 10 $$
?
 A: HINT:
$$\begin{align}10^{20000}=&(10^{100}+3)\cdot\bigl(10^{19900}-3\cdot 10^{19800}+9\cdot10^{19700}-27\cdot 10^{19600}\pm\ldots\\&+(-3)^{k}\cdot 10^{19900-100k}\pm\cdots+(-3)^{199}+\epsilon\bigr)\end{align}$$ with $\epsilon\approx 3^{100}10^{-100}>0$
A: 
This didn't work as a comment to Hagen von Eitzen's answer, so I post it here.

Writing the fraction as
$$
\small\begin{align}
&\frac{10^{20000}}{10^{100}}\frac1{1+3\cdot10^{-100}}\\
&=10^{19900}\left(1-3\cdot10^{-100}+\left(3\cdot10^{-100}\right)^2-\dots
\color{#00A000}{-\left(3\cdot10^{-100}\right)^{199}}\color{#C00000}{+\left(3\cdot10^{-100}\right)^{200}-\dots}\right)\\
&\equiv\color{#00A000}{-3^{199}}\color{#C00000}{+\epsilon}\pmod{10^{100}}
\end{align}
$$
where $0\le\epsilon\le\left(\frac{9}{10}\right)^{100}$
Thus,
$$
\begin{align}
\left\lfloor\frac{10^{20000}}{10^{100}+3}\right\rfloor
&\equiv-3^{199}\pmod{10^{100}}\\
&\equiv-\left(3^4\right)^{49}\,3^3\pmod{10}\\[6pt]
&\equiv-1^{49}\,7\pmod{10}\\[12pt]
&\equiv3\pmod{10}
\end{align}
$$
