Competition problem — Number theory

Library / /21 of 224

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

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.