3. How many solutions does the equation

$$
x-2019\{x\}=2019
$$

have in the set of real numbers? (For a real number $x$, $\lfloor x\rfloor$ denotes the greatest integer not greater than $x$, and $\{x\}$ denotes the value $x-\lfloor x\rfloor$.)
3. Given $\{x\} \in [0,1)$, it is clear that all solutions $x$ can only be in the interval [2019, 4038). Let's introduce the substitution $x = 2019 + t$. Then $\{x\} = \{t\}$, and the initial equation reduces to $t = 2019\{t\}$, where $t \in [0, 2019)$. From the expression $t = \lfloor t \rfloor + \{t\}$, we get

![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-37.jpg?height=451&width=485&top_left_y=808&top_left_x=1031)

For 2019 1A 2 $\lfloor t \rfloor = 2018\{t\}$, i.e., $\{t\} = \frac{\lfloor t \rfloor}{2018}$. Since $\lfloor t \rfloor$ can take values $0, 1, 2, \ldots, 2018$,

![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-37.jpg?height=712&width=142&top_left_y=1384&top_left_x=244)

For 2019 1A 4, for each of these possibilities, we will get a unique value of $\{t\}$, with the exception of the case $\lfloor t \rfloor = 2018$: in this case, it would follow that $\{t\} = \frac{2018}{2018} = 1$, which is impossible due to the requirement $0 \leq \{t\} < 1$. Therefore, we conclude that the obtained equation has a total of 2018 solutions for $t$ (and these are: $0, 1 + \frac{1}{2018}, 2 + \frac{2}{2018}, 3 + \frac{3}{2018}$, $\left.2017 + \frac{2017}{2018}\right)$, and thus the original equation has a total of 2018 solutions.