How to solve this equation involving the floor function? How to solve  the equation
$$\left[ \frac x {1!} \right]  +  \left[ \frac x {2!} \right] +\left[ \frac x {3!} \right]+\cdots +  \left[ \frac x {10!} \right]  =1001,$$
where $[ r]$ stands for the integer part of a real number $r,$
in the integers? I can do it with Maple only.
 A: Using inequalities reduce the problem, as
$$
y-1< [y] \le y
$$
we have
$$
\left(\frac{x}{1!}-1\right)+\cdots+\left(\frac{x}{10!}-1\right) < 1001 \le \frac{x}{1!}+\cdots+\frac{x}{10!}
$$
simplifying into
$$
x\left(\frac{1}{1!}+\cdots+\frac{1}{10!}\right)-10 < 1001 \le x\left(\frac{1}{1!}+\cdots+\frac{1}{10!}\right)
$$
solving the two inequalities in $x$ gives (its just computation)
$$
582\le x \le 589
$$
Then, I don't really know how to do... trying all values or doing bissection can be the only way to solve this, as floor function is a bit random...
Trying all integers between 582 and 589 we find the solution $x=584$ as previously said.
A: Let's play with
$$f(x)=\sum_{r=1}^{10}\left[\frac{x}{r!}\right]$$
$f(90)=153$
$f(100)=170$
$f(150)=257$
$f(500)=857$
I found a pattern. $\frac{153}{90}\approx\frac{170}{100}\approx\frac{257}{170}\approx\frac{857}{500}\approx 1.7$
With higher numbers, error is of max $ 0.02$
Hence, we should start guessing our number near $\frac{1001}{1.7}=588.82$
$f(590)=1011$
$f(580)=944$
Calculation must not be tough now. Just see how which fraction changes when you increase to $581$
$f(581)=995$
$f(582)=998$
$f(583)=999$
$f(584)=1001$
and we are done. 
We also could have used $1.71$ which is better approximation with large numbers.
This yield initial guess of $585.38$. More closer.
This is related with $$\lim_{x\to\infty}\dfrac{f(x)}{x}$$
My calculator plots a somewhat converging graph. 
Let me see if I can find the limit. It seems to be $e-1$
A: Let's find $\lim_{x\to\infty}\dfrac{f(x)}{x}$ where $f(x)=\sum_{r=1}^{10}\left[\frac{x}{r!}\right]$
Note that you can write numerator as $\sum\dfrac{x}{k!} + \text{error}$
Note that this $\text{error}$ is not more that $10$ and hence we have denominator as $x$ which is very large as it tends to infinity and it will make it zero.
We also note that $e-1=\sum_{k\ge1}\dfrac{1}{k!}$
Hence, limit is $\sum_{k=1}^{10}\dfrac{1}{k!}\approx e-1$
Now you can start you initial guess near $\dfrac{1001}{e-1}\approx \dfrac{1001}{1.718}\approx 582.65$
This gives us integer $x=584$
Also $f(x+r)=f(x) \forall r\in[0,1)\implies x\in[584,585) $
