China Southeastern Mathematical Olympiad

First, we point out two obvious facts: (a) If $m$ is a positive integer and $x$ is real, then $$\lfloor \frac{x}{m}\rfloor =\lfloor \frac{\lfloor x\rfloor }{m}\rfloor .$$ (b) For any integer $l$ and positive even number $m$, we have $$\lfloor \frac{2l+1}{m}\rfloor =\lfloor \frac{2l}{m}\rfloor .$$ Let $m=k!$ ($k=1,2,\dots ,2013$) in (a) and summing up, we have $$f(x)=\sum _{k=1}^{2013}\lfloor \frac{x}{k!}\rfloor =\sum _{k=1}^{2013}\lfloor \frac{\lfloor x\rfloor }{k!}\rfloor =f(\lfloor x\rfloor ),$$ that is, $f(x)=n$ has a real solution if and only if $f(x)=n$ has an integer solution. So, we only consider $x$ as an integer. Since $$f(x+1)-f(x)=\lfloor x+1\rfloor -\lfloor x\rfloor +\sum _{k=2}^{2013}\left(\lfloor \frac{x+1}{k!}\rfloor -\lfloor \frac{x}{k!}\rfloor \right)\ge 1,\stackrel{◯}{1}$$ we see that $f(x)$ ($x\in \mathbb{Z}$) is monotonously increasing. Now, we find integers $a$ and $b$, such that $$\begin{array}{rl}f(a-1)<0& \le f(a)<f(a+1)<\dots \\ & <f(b-1)<f(b)\le 2013<f(b+1).\end{array}$$ Note that $f(-1)<0=f(0)$, so $a=0$. Since $$\begin{array}{rl}f(1173)& =\sum _{k=1}^6\lfloor \frac{1173}{k!}\rfloor =1173+586+195+48+9+1\\ & =2012\le 2013,\end{array}$$ $$\begin{array}{rl}f(1174)& =\sum _{k=1}^6\lfloor \frac{1174}{k!}\rfloor =1174+587+195+48+9+1\\ & =2014>2013,\end{array}$$ we see that $b=1173$. So the good numbers in $\{1,3,5,...,2013\}$ are the odd numbers in $$\{f(0),f(1),\dots ,f(1173)\}.$$ Let $x=2l$ ($l=0,1,\dots ,586$) in ①. By (b), we have $$\lfloor \frac{2l+1}{k!}\rfloor =\lfloor \frac{2l}{k!}\rfloor (2\le k\le 2013).$$ Thus, $$f(2l+1)-f(2l)=1+\sum _{k=2}^{2013}\left(\lfloor \frac{2l+1}{k!}\rfloor -\lfloor \frac{2l}{k!}\rfloor \right)=1,$$ that is, there is exactly one odd number in $f(2l)$ and $f(2l+1)$. Therefore, there are $\frac{1174}{2}=587$ odd numbers in $\{f(0),f(1),\dots ,f(1173)\}$, that is, there are 587 good numbers in the set $\{1,3,5,\dots ,2013\}$. $\square$