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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

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$2009\equiv -2(\mod 2011), 2010\equiv -1(\mod 2011)$ –  Arturo Dec 8 '13 at 10:25
    
We can replace $2011$ with $n$ such that $\displaystyle \frac{n-1}2$ is odd $\iff\displaystyle n\equiv-1\pmod4$ –  lab bhattacharjee Dec 8 '13 at 12:22

2 Answers 2

up vote 4 down vote accepted

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 $0 \bmod 11$.

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Oops, I got my signs mixed up, I will fix this and hopefully make it clearer. –  fretty Dec 8 '13 at 11:02
    
Looks better now. –  Marc van Leeuwen Dec 8 '13 at 11:11

Multiply every factor in the first product by $-1$ to give, modulo$~2011$, the mirror image factor of the second product ($-1\times1\equiv2010$, $-1\times3\equiv2008$, etc.). That gives $2010/2=1005$ factors$~{-}1$ in all, whose product is$~{-}1$. Hence the second product is the opposite of the first, of course still modulo$~2011$.

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