Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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

share|cite|improve this question
$2009\equiv -2(\mod 2011), 2010\equiv -1(\mod 2011)$ – 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
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 $0 \bmod 11$.

share|cite|improve this answer
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$~2011$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.