Demetres wrote: Determine all real numbers $x\in\mathbb{R}$ for which \[ \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor=x-2022 \] $LHS\in\mathbb Z$ and so $RHS\in\mathbb Z$ and so $x\in\mathbb Z$ If $x=6n$, equation is $3n+2n=6n-2022$ and so $n=2022$ and $x=12132$ If $x=6n+1$, equation is $3n+2n=6n+1-2022$ and so $n=2021$ and $x=12127$ If $x=6n+2$, equation is $3n+1+2n=6n+2-2022$ and so $n=2021$ and $x=12128$ If $x=6n+3$, equation is $3n+1+2n+1=6n+3-2022$ and so $n=2021$ and $x=12129$ If $x=6n+4$, equation is $3n+2+2n+1=6n+4-2022$ and so $n=2021$ and $x=12130$ If $x=6n+5$, equation is $3n+2+2n+1=6n+5-2022$ and so $n=2020$ and $x=12125$ Anbd so $\boxed{x\in\{12125,12127,12128,12129,12130,12132\}}$