How many numbers from $1-1000$ can be written in the form $$⌊2x⌋+⌊4x⌋+⌊6x⌋+⌊8x⌋$$ where $\left\lfloor\cdot\right\rfloor$ denotes the floor function?
This is an AIME 1985 problem. I am confused by its solution. In the solution, it says:
The value $$\{2x\}+\{4x\}+\{6x\}+\{8x\}$$ changes only when $x=\frac{m}{n}$ where $m=\{1,2,3,..,n-1\}$ and $n=\{2,4,6,8\}$.
I don't understand why this will be true. Can anyone give me any proof?
For integers $n$ and $m$, $floor(nx) = m$ means $m \le nx < m+1$, i.e. (if $n > 0$) $\frac{m}{n} \le x < \frac{m+1}{n}$. Thus $floor(nx)$ changes at $x = m/n$ where $m$ is an integer. The sum of several of these terms can only change when one or more of the terms change. The statement "$m = \{1,2,3,\ldots,n-1\}$" is wrong though. Instead, you want $1 < x = m/n < 1000$, so $n < m < 1000 n$.