6. Given $[x]$ represents the greatest integer not exceeding the real number $x$. If $P=\sum_{i=1}^{2010}\left[\frac{6^{i}}{7}\right]$, then the remainder when $P$ is divided by 35 is
6. 20 .

First, consider $S=\frac{6}{7}+\frac{6^{2}}{7}+\cdots+\frac{6^{2010}}{7}$.
In equation (1), no term is an integer, but the sum of any two adjacent terms is an integer (since $\frac{6^{k}}{7}+\frac{6^{k+1}}{7}=6^{k}$ $(k \in \mathbf{Z})$ is an integer).

If the sum of two non-integer numbers is an integer, then the sum of their integer parts is 1 less than the sum of the numbers themselves. Since there are 2010 terms in equation (1), the value of the desired sum $P$ is 1005 less than $S$.
Using the formula for the sum of a geometric series, we get
$$
\begin{array}{l}
S=\frac{6\left(6^{2010}-1\right)}{35} . \\
\text { Therefore, } 35 P=35(S-1005) \\
=6^{2011}-6-35 \times 1005 . \\
\text { Also, } 6^{2011}=6(35+1)^{1005} \\
=6 \times 35^{2} M+6 \times 35 \times 1005+6\left(M \in \mathbf{N}_{+}\right) \\
\Rightarrow 35 P=6 \times 35^{2} M+5 \times 35 \times 1005 \\
\Rightarrow P=6 \times 35 M+5 \times 1005 .
\end{array}
$$

Therefore, the remainder when $P$ is divided by 35 is equal to
$$
5 \times 1005=5025
$$

divided by 35.
Thus, the remainder when $P$ is divided by 35 is 20.