6. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If
$$
A=\left[\frac{7}{8}\right]+\left[\frac{7^{2}}{8}\right]+\cdots+\left[\frac{7^{2019}}{8}\right]+\left[\frac{7^{2020}}{8}\right],
$$

then the remainder when $A$ is divided by 50 is
6. 40.

Notice that, $\frac{7^{2 k-1}}{8}$ and $\frac{7^{2 k}}{8}$ are not integers.
By definition,
$$
\begin{array}{l}
7^{2 k-1}-2=\left(\frac{7^{2 k-1}}{8}-1\right)+\left(\frac{7^{2 k}}{8}-1\right) \\
<\left[\frac{7^{2 k-1}}{8}\right]+\left[\frac{7^{2 k}}{8}\right]<\frac{7^{2 k-1}}{8}+\frac{7^{2 k}}{8} \\
=7^{2 k-1}.
\end{array}
$$

Thus, for any positive integer $k$, we have
$$
\begin{array}{l}
{\left[\frac{7^{2 k-1}}{8}\right]+\left[\frac{7^{2 k}}{8}\right]=7^{2 k-1}-1=7 \times 7^{2 k-2}-1} \\
=7 \times 49^{k-1}-1=7 \times(50-1)^{k-1}-1 \\
\equiv 7 \times(-1)^{k-1}-1(\bmod 50). \\
\text{Therefore, } A \equiv 7 \sum_{k=1}^{1010}(-1)^{k-1}-1010 \\
\equiv 40(\bmod 50).
\end{array}
$$