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Belarusian Mathematical Olympiad

Number theory Difficulty 5.1 AIME, harder Prove it Belarus

Find $\left\{ \frac{2009!}{2011!} \right\}$ . (Here $\{x\}$ means the fractional part of $x$ .)

Solution

Answer: $\frac{1}{2011}$ .
Since $2011$ is a prime number, we see that $2010! \equiv -1 \pmod{2011}$ (Wilson's theorem). Hence $2010! \equiv 2010 \pmod{2011}$ , which gives

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