Library / /431 of 740

, 2023

Number theory Difficulty 5.0 AIME, harder Prove it United States

Problem:
An integer $n$ is chosen uniformly at random from the set $\{1,2,3, \ldots, 2023!\}$ . Compute the probability that
$\gcd\left(n^{n}+50, n+1\right)=1$

Solution

Solution:
If $n$ is even, we need $\gcd(n+1,51)=1$ . If $n$ is odd, we need $\gcd(n+1,49)=1$ . Thus, the answer is
$\frac{1}{2}\left(\frac{\varphi(49)}{49}+\frac{\varphi(51)}{51}\right)=\frac{265}{357}$

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.