6. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $A=\left[\frac{7}{8}\right]+\left[\frac{7^{2}}{8}\right]+\left[\frac{7^{2019}}{8}\right]+\left[\frac{7^{2020}}{8}\right]$, then the remainder when $A$ is divided by 50 is $\qquad$ .
Since $\frac{7^{2 k-1}}{8}+\frac{7^{2 k}}{8}=7^{2 k-1} \Rightarrow\left[\frac{7^{2 k-1}}{8}\right]+\left[\frac{7^{2 k}}{8}\right]=7^{2 k-1}-1$,
thus $A=7+7^{3}+\cdots+7^{2019}-1010=7\left(1+7^{2}+\cdots+7^{2018}\right)-1010$, since the number inside the parentheses is divisible by 50, so $A \equiv-1010 \equiv 40(\bmod 50)$.