13. Find the remainder when $(1!\times 1)+(2!\times 2)+(3!\times 3)+\cdots+(286!\times 286)$ is divided by 2009 .
13. Answer: 2008
Observe that $n!\times n=n!\times(n+1-1)=(n+1)!-n!$. Therefore
$$
\begin{aligned}
& (1!\times 1)+(2!\times 2)+(3!\times 3)+\cdots+(286!\times 286) \\
= & (2!-1!)+(3!-2!)+(4!-3!)+\cdots+(287!-286!) \\
= & 287!-1 .
\end{aligned}
$$

Since $2009=287 \times 7,287!-1=-1=2008(\bmod 2009)$. It follows that the remainder is 2008 .