Calculate $2008!\pmod {2011}$

Question

I think you should use the theorem of Wilson, because 2011 is a prime number. But I don't know how to use it. Thanks

Answer 1

Well, you're right, this can be done fairly easily using Wilson's theorem. Since $2011$ is prime, by Wilson's theorem, $$2010!\equiv -1\pmod{2011}$$

Now let $k=2008!$. Then

$$-1\equiv2010!\equiv(2010)(2009)k\equiv(-1)(-2)k\equiv 2k\pmod{2011}$$

So we get $$-1\equiv 2k\pmod{2011}$$

which implies $k\equiv1005\pmod{2011}$.

Answer 2

$$(2010)!\equiv-1\pmod{2011}$$

Now $(2010)!=2008!\cdot2009\cdot2010\equiv2008!(-2)(-1)2\cdot2008!$

$2008!\equiv2^{-1}\equiv1006$ as $2\cdot1006\equiv1\pmod{2011}$

Answer 3

Hint: Wilson's theorem will certainly be helpful here. $$2010!\equiv -1\pmod{2011}$$ $$2009!\equiv 1\pmod{2011}$$ $$2008!\equiv 1005\pmod{2011}$$

Answer 4

You're right about the approach you mention.

From Wilson's theorem it follows that $2008!\cdot2009\cdot2010 = 2010!\equiv -1\mod 2011$. Moreover, since $2011$ is a prime number inverses exist, so $2008! = 2008!\cdot2009\cdot2010\cdot 2009^{-1}\cdot2010^{-1} = 2010!\cdot 2009^{-1}\cdot2010^{-1} = - 2009^{-1}\cdot2010^{-1}$.

Since $2010\equiv-1$, it follows that $2010^2=(-1)^2=1$, so $2010$ is it's own inverse. We also know that $2009\equiv-2$, while $-2\cdot-1006=2012\equiv 1$. Hence $2009^{-1}\equiv -1006=1005$.

Combining all of this, we find that $2008! \equiv -1\cdot-1\cdot1005 = 1005$.

Answer 5

Wilson's Theorem states that $(p-1)! \equiv -1 \pmod p$ for any prime number $p$. Thus, we know that $2010! \equiv -1\equiv 2010 \pmod{2011}$, and that $2009! \equiv 1 \pmod{2011}$. However, since $2009 \equiv -2 \pmod{2011}$ we have $(-2)(2008!)\equiv 1 \pmod {2011}$ and so $2008! \equiv 1005 \pmod{2011}$.