Question 67
Let $x_{1}, x_{2}, \ldots, x_{2022}$ be nonzero real numbers. Suppose that $x_{k}+\frac{1}{x_{k+1}}\lt 0$ for each $1 \leq k \leq 2022$, where $x_{2023}=x_{1}$. Compute the maximum possible number of integers $1 \leq n \leq 2022$ such that $x_{n}\gt 0$.
Proposed by: Akash Das
Proposed by: Akash Das