1. (10 points) Let $[x]$ denote the greatest integer not exceeding $x$, for example $[3.14]=3$, then $\left[\frac{2017 \times 3}{11}\right]+\left[\frac{2017 \times 4}{11}\right]+\left[\frac{2017 \times 5}{11}\right]+\left[\frac{2017 \times 6}{11}\right]+\left[\frac{2017 \times 7}{11}\right]+\left[\frac{2017 \times 8}{11}\right]$ is . $\qquad$
【Analysis】We can first simplify the original expression by converting each term into a mixed number form, then take the integer part of each, which will give us the sum.
【Solution】According to the analysis, the original expression is:
$$
\begin{array}{l}
{\left[\frac{2017 \times 3}{11}\right]+\left[\frac{2017 \times 4}{11}\right]+\left[\frac{2017 \times 5}{11}\right]+\left[\frac{2017 \times 6}{11}\right]+\left[\frac{2017 \times 7}{11}\right]+\left[\frac{2017 \times 8}{11}\right]} \\
=\left[550 \frac{1}{11}\right]+\left[733 \frac{5}{11}\right]+\left[916 \frac{9}{11}\right]+\left[1100 \frac{2}{11}\right]+\left[1283 \frac{6}{11}\right]+\left[1466 \frac{10}{11}\right] \\
=550+733+916+1100+1283+1466 \\
=6048 .
\end{array}
$$

Therefore, the answer is 6048.