jasperE3 wrote: Find all positive numbers $x$ such that $20\{x\}+0.5\lfloor x\rfloor = 2005$. Let $x=n+y$ with $n\in\mathbb Z$ and $y\in[0,1)$ Equation is $20y+\frac n2=2005$ and so $y=\frac{4010-n}{40}$ and $x=\frac{4010+39n}{40}$ Constraint $y\in[0,1)$ easily gives $n\in[3971,4010]$ Hence the answer $\boxed{x=\frac{4010+39n}{40}\text{, whatever is integer }n\in[3971,4010]}$