How prove this $1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots 2010\equiv 0 (\mod2011)$ Question:
show that
$$1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots 2010\equiv 0 (\mod2011)$$
my try: since
\begin{align*}&1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots \times2010\\
&=(2009)!!+2^{1005}\cdot1005!!
\end{align*}
then I can't.Thank you for you help

Ok so for the first term:
$1\times 3 \times 5 \times ... \times 2009$ 
$\equiv 1 \times 3 \times 5 \times ... \times 1005 \times (-1004) \times... \times (-4) \times (-2)$ 
$\equiv (1 \times 2 \times ... \times 1005)$ 
$\equiv 1005! \bmod 2011.$
The second term is similar:
$2\times 4\times 6 \times ... \times 2010$
$\equiv 2 \times 4 \times 6 \times ... \times 1004 \times (-1005) \times ... \times (-3) \times (-1)$ 
$\equiv -(1 \times 2 \times ... \times 1005)$
$\equiv -1005! \bmod 2011.$
So the sum is $0 \bmod 11$.
