9. $A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{0114}}{9}\right]$ when divided by 63 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
9. 56 .

Notice that, for any positive integer $k$, $\frac{8^{2 k-1}}{9}$ and $\frac{8^{2 k}}{9}$ are not integers, and
$$
\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}=8^{2 k-1} \text {. }
$$

Therefore, for any positive integer $k$, we have
$$
\begin{array}{l}
{\left[\frac{8^{2 k-1}}{9}\right]+\left[\frac{8^{2 k}}{9}\right]=\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}-1} \\
=8^{2 k-1}-1 \equiv 7(\bmod 63) . \\
\text { Hence } A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{2014}}{9}\right] \\
\equiv 1007 \times 7 \equiv 56(\bmod 63) .
\end{array}
$$