1. Let $[x]$ be the greatest integer not greater than $x$, and let $\{x\}=x-[x]$. Find all positive real numbers $x$ for which

$$
20\{x\}+0.5[x]=2005
$$
IV/1. Multiply the equation by 2. Then we can write it as $40\{x\}=4010-[x]=a$, where $a$ is some integer. Therefore, $\{x\}=\frac{a}{40}$ for $0 \leq a \leq 39$. We have $[x]=4010-a$, so we get 40 solutions of the form $x=4010-a+\frac{a}{40}$, where $a=0,1, \ldots, 39$.