6. Define the sequence $\left\{a_{n}\right\}: a_{n}$ is the last digit of $1+2+\cdots+n$, and $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then $S_{2016}=$ $\qquad$
6. 7066 .

From the problem, we know
$$
\begin{array}{l}
\frac{(n+20)(n+20+1)}{2}=\frac{n^{2}+41 n+420}{2} \\
=\frac{n(n+1)}{2}+20 n+210 .
\end{array}
$$

Then $\frac{(n+20)(n+21)}{2}$ and $\frac{n(n+1)}{2}$ have the same last digit, i.e., $a_{n+20}=a_{n}$.
$$
\begin{array}{l}
\text { Thus, } S_{2016}=S_{16}+100 S_{20} . \\
\text { Also, } S_{20}=a_{1}+a_{2}+\cdots+a_{20} \\
=1+3+6+0+5+1+8+6+5+5+ \\
\\
6+8+1+5+0+6+3+1+0+0 \\
=70, \\
S_{16}=66,
\end{array}
$$
$$
\text { Therefore, } S_{2016}=S_{16}+100 S_{20}=7066 \text {. }
$$