Rightmost digit of $ \left \lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor $ How could I find $$ 0 \leq a \leq 9 $$ such that
$$ \left \lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor \equiv a \mod 10 $$
?

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 $\epsilon\approx 3^{100}10^{-100}>0$
