question
Solve in the set of real numbers the equation $\lfloor 2022 x \rfloor+\{2023 x\}=2023$, where (where $\lfloor a\rfloor$ and $\{a\}$ represent the whole part, respectively the fractional part of the real number $a$)
my idea
Can you tell me if my idea is right or wrong? If so, can you tell me how i should solve this problem.
First of all i wrote $\lfloor 2022 x\rfloor +\{2023 x\}=2023$ as $2022x-\{2022 x\}+\{2023 x\}=2023$ which means that $\{2023 x\}-\{2022 x\}=2023-2022x$
We know that $-1<{2023 x}-{2022 x}<1$ which means that $-1<2023-2022 x<1$
From this we get that $1<x<\frac{2024}{2022}$
Hope one of you can help me! Thank you!
Best Answer
Your considerations are correct, but incomplete. Relying on rote algebraic manipulations is not going to be particularly useful here. Instead, start by thinking about what obvious properties the two summands have:
The combination of 2 and 3 tells us that $\{2023x\} = 0$, i.e. that $2023x$ is itself a natural number, and that $\lfloor 2022x\rfloor = 2023$.
Since $2023x \in \mathbb{N}$, we know that $x = \frac{n}{2023}$ for some $n \in \mathbb{N}$. The remaining requirement is that $\lfloor \frac{2022n}{2023}\rfloor = 2023$. Clearly, this requires $n > 2023$. We can then test the following values by numerical calculation. We find that $2024$ is still a little too small. Taking $n = 2025$ works. And then $2026$ is already too large.
Thus, the unique solution to $\lceil 2022x\rceil + \{2023x\} = 2023$ is $x = \frac{2025}{2023}$.