$\lfloor x\rfloor+\frac{1999}{\lfloor x\rfloor}=\{x\}+\frac{1999}{\{x\}} \Longrightarrow \lfloor x\rfloor - \{x\} = 1999(\frac{\lfloor x\rfloor - \{ x \}}{\lfloor x\rfloor \cdot \{x\}}) \Longrightarrow \lfloor x\rfloor \cdot \{ x \} = 1999 \Longrightarrow \{x \} = \frac{1999}{\lfloor x\rfloor}$. Since $0 \leq \{ x \} \le 1$, we have $\lfloor x\rfloor \geq 2000$. Now let $\lfloor x\rfloor = n$ where $n \geq 2000$, then $\{ x\} = \frac{1999}{n}$. So all solutions are $\boxed{x = n + \frac{1999}{n}}$, where $n \geq 2000$ and $n \in \mathbb{N}$.

jasperE3 wrote: Solve the equation $\lfloor x\rfloor+\frac{1999}{\lfloor x\rfloor}=\{x\}+\frac{1999}{\{x\}}$. This suitable for HSM. Edit-: Now Posted in HSM from HSO

My solution will be same as other solutions :