s372102 wrote: $f(x)=\sum\limits_{i=1}^{2013}\left[\dfrac{x}{i!}\right]$. A integer $n$ is called good if $f(x)=n$ has real root. How many good numbers are in $\{1,3,5,\dotsc,2013\}$? there are some minor errors in the statement. Here an improved version: Let $f(x)=\sum\limits_{i=1}^{2013}\left[\dfrac{x}{i!}\right]$. Integer $n$ is called a "good number" if equation $f(x)=n$ has at least a real solution. How many "good numbers" are there in the set $\{1,3,5,\dotsc,2013\}$?

s372102 wrote: $f(x)=\sum\limits_{i=1}^{2013}\left[\dfrac{x}{i!}\right]$. A integer $n$ is called good if $f(x)=n$ has real root. How many good numbers are in $\{1,3,5,\dotsc,2013\}$? $f(x)$ is increasing and $\forall x\in [1,7!)$ : $f(x)=x+\left\lfloor\frac x2\right\rfloor$ $+\left\lfloor\frac x6\right\rfloor$ $+\left\lfloor\frac x{24}\right\rfloor$ $+\left\lfloor\frac x{120}\right\rfloor$ $+\left\lfloor\frac x{720}\right\rfloor$ $1+\frac 12+\frac 16+\frac 1{24}$ $+\frac 1{120}+\frac 1{720}$ $=\frac {1237}{720}$ So $\forall x\in[1,5040)$ : $5+\frac {1237}{720}x\ge f(x)> \frac {1237}{720}x-5$ and $\forall x\ge 5040$ $f(x)>x\ge 5040$ So $f(1171)<2013$ and $f(1175)>2013$ and very easy quick calculus gives $f(1172)=2011$ and $f(1173)=2012$ and $f(1174)=2014$ So we have exactly $1173$ good numbers in $\{1,2,3,4,...,2012,2013\}$ It remains to deal with odd and even such numbers. $\forall x\in[1,720]$ : $f(x+720)=f(x)+1237$ and so we have exactly $720$ odd numbers and $720$ even numbers in $f([1,1440])$ $f(1440)=2474$ is even $\forall x\in[1416,1439]$ : $f(x)-24=f(x)-41$ and so we have exacty $24$ odd numbers and $24$ even numbers in $f([1392,1439])$ $\forall x\in[1368,1391]$ : $f(x)-24=f(x)-41$ and so we have exacty $24$ odd numbers and $24$ even numbers in $f([1344,1391])$ $f(1338)=2297$ $f(1339)=2298$ $f(1340)=2300$ $f(1341)=2301$ $f(1342)=2303$ $f(1343)=2304$ And so we have exactly $3$ odd numbers and $3$ even numbers in $f([1338,1343])$ $\forall x\in[1338,1343]$ : $f(x)-6=f(x)-10$ and so we have exacty $3$ odd numbers and $3$ even numbers in $f([1332,1337])$ $\forall x\in[1332,1337]$ : $f(x)-6=f(x)-10$ and so we have exacty $3$ odd numbers and $3$ even numbers in $f([1326,1331])$ $f(1320)=2267$ $f(1321)=2268$ $f(1322)=2270$ $f(1323)=2271$ $f(1324)=2273$ $f(1325)=2274$ And so we have exactly $3$ odd numbers and $3$ even numbers in $f([1320,1325])$ $\forall x\in[1296,1319]$ : $f(x)-24=f(x)-41$ and so we have exacty $24$ odd numbers and $24$ even numbers in $f([1272,1319])$ $\forall x\in[1248,1271]$ : $f(x)-24=f(x)-41$ and so we have exacty $24$ odd numbers and $24$ even numbers in $f([1224,1271])$ Using previous lines, we have exactly $12$ odd numbers and $12$ even numbers in $f([1320,1343])$ $\forall x\in [1320,1343]$ : $f(x-120)=f(x)-206$ and so we have exactly $12$ odd numbers and $12$ even numbers in $f([1200,1223])$ So we have exactly $600$ odd numbers and $599$ even numbers in $f([1,1199])$ $f(1194)=2049$ $f(1195)=2050$ $f(1196)=2052$ $f(1197)=2053$ $f(1198)=2055$ $f(1199)=2056$ And so we have exactly $3$ odd numbers and $3$ even numbers in $f([1194,1199])$ $\forall x\in[1194,1199]$ : $f(x)-6=f(x)-10$ and so we have exacty $3$ odd numbers and $3$ even numbers in $f([1188,1193])$ $\forall x\in[1188,1193]$ : $f(x)-6=f(x)-10$ and so we have exacty $3$ odd numbers and $3$ even numbers in $f([1182,1187])$ $\forall x\in[1182,1187]$ : $f(x)-6=f(x)-10$ and so we have exacty $3$ odd numbers and $3$ even numbers in $f([1176,1181])$ $f(1174)=2014$ $f(1175)=2015$ So we have exactly $13$ odd numbers and $13$ even numbers in $f([1174,1199])$ So we have exactly $587$ odd numbers and $586$ even numbers in $f([1,1173])$ Hence the answer : $\boxed{587}$

CSMO 2013 wrote: $f(x)=\sum\limits_{i=1}^{2013}\left[\dfrac{x}{i!}\right]$. A integer $n$ is called good if $f(x)=n$ has real root. How many good numbers are in $\{1,3,5,\dotsc,2013\}$? Here's a shorter proof to the answer $587$. In general, not a bad problem. Notice that for any $x\in \mathbb R$, $f(x)=f([x])$, so $x$ is a root of $f(x)-n$ iff it's an integer. Now, observe that $f$ is increasing and $1=f(1)<\dots<f(1173)=2012<2013<f(1174)$ (thanks to @pco here). So we only need to determine the number of odd numbers among $\{f(1),f(2), \dots, f(1173)\}$. It's easy to see that if $\alpha\in \mathbb Z$, $f(\alpha+1)$ is $f(\alpha)+1$ and $f(\alpha)+2$ if $\alpha$ is even and odd respectively. So the arrow of odd values goes as follows (assuming $f(1)$ is odd) : $$f(1)\to f(2)\to f(5)\to f(6)\to \dots \to f(1169)\to f(1170)\to f(1173)$$, a total of $587$ numbers.