5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set
$$
A=\left[\frac{19}{20}\right]+\left[\frac{19^{2}}{20}\right]+\cdots+\left[\frac{19^{2020}}{20}\right] .
$$

Then the remainder when $A$ is divided by 11 is $\qquad$
5.2.

Notice that, $\left[\frac{19^{2 n-1}}{20}\right] 、\left[\frac{19^{2 n}}{20}\right]$ are not integers and
$$
\frac{19^{2 n-1}}{20}+\frac{19^{2 n}}{20}=19^{2 n-1} \text {. }
$$

Therefore, for any positive integer $n$, we have
$$
\begin{array}{l}
{\left[\frac{19^{2 n-1}}{20}\right]+\left[\frac{19^{2 n}}{20}\right]=19^{2 n-1}-1} \\
\equiv(-3)^{2 n-1}-1(\bmod 11) .
\end{array}
$$

Then, $A=19^{1}+19^{3}+\cdots+19^{2019}-1010$
$$
\begin{array}{l}
\equiv(-3)^{1}+(-3)^{3}+\cdots+(-3)^{2019}-1010 \\
\equiv \frac{-3\left(9^{1010}-1\right)}{9-1}-1010 \\
\equiv \frac{-3\left(2^{1010}-1\right)}{8}-1010 \\
\equiv \frac{-3\left(32^{202}-1\right)}{8}-1010 \\
\equiv \frac{-3\left((-1)^{202}-1\right)}{8}-1010 \\
\equiv 2(\bmod 11)
\end{array}
$$