How does one find all integer solutions to this equation: $$ \biggl\lfloor{\frac{x}{1!}\biggr\rfloor} + \biggl\lfloor{\frac{x}{2!}\biggr\rfloor} + \cdots +\biggl\lfloor{\frac{x}{10!}\biggr\rfloor}=1001$$
Can't think about methods for solving this. Though i did something i am not sure of it.
The left hand side is strictly less than $(e-1)x$ and so you can start testing at $x=\biggl\lceil{\dfrac{1001}{e-1}}\biggr\rceil = 583$ and continue until you get a sum larger than 1001, something which will happened very quickly; in this particular case $x=585$ produces a sum which is too large.
I noticed that $x=6!$ is too big. So the solution solves:
$$\biggl\lfloor{\frac{x}{1!}\biggr\rfloor} + \biggl\lfloor{\frac{x}{2!}\biggr\rfloor} + \cdots \biggl\lfloor{\frac{x}{5!}\biggr\rfloor}=1001.$$
Then I divided $\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} +\frac{1}{4!} + \frac{1}{5!}$ into $1001$ and got $583.107$. Trying $584$ in the equation works. Any integer larger or smaller will not solve the equation, but $584 \leq x < 585$.
We are given $\lfloor x/1!\rfloor + \lfloor x/2! \rfloor + \dots + \lfloor x/10!\rfloor = 1001$.
Now,
$$\begin{align} \lfloor x/1!\rfloor \leq 1001 &\Longrightarrow x \leq 1001 \\ &\Longrightarrow x/7! \leq 1001/7! \\ &\Longrightarrow x/7! \leq 1001/5040 \\ &\Longrightarrow \lfloor x/7! \rfloor = 0 \\ &\Longrightarrow \lfloor x/k! \rfloor = 0, \quad\text{ for } k \in\{ 7,8,9,10 \} \end{align}$$
It suffices to solve $\lfloor x/1! \rfloor + \dots + \lfloor x/6! \rfloor = 1001$. Since $y - 1 < \lfloor y \rfloor \leq y$, we have
$$\begin{align} x(1/1! + \dots 1/6!) - 6 < \lfloor x/1!\rfloor + \dots + \lfloor x/6! \rfloor \leq x(1/1! + \dots + 1/6!) \end{align}$$
and thus $582.635 \leq x < 585.546$. Among the integers $583, 584$, and $585$, the integer $584$ satisfies the equation.
Every non-negative integer can be written (uniquely) in the form
$$x=a_1+2a_2+6a_3+24a_4+120a_5+\cdots$$
where $0\le a_k\le k$. It's easy to show that
$$\big\lfloor {x\over1}\big\rfloor+\big\lfloor {x\over2}\big\rfloor+\big\lfloor {x\over6}\big\rfloor+\big\lfloor {x\over24}\big\rfloor+\cdots=a_1+3a_2+10a_3+41a_4+206a_5+1237a_6+\cdots$$
where the sequence of coefficients $c_k=1,3,10,41,206,1237,\ldots$ satisfies the recursion $c_k=kc_{k-1}+1$. If $1001$ can be represented by such a sum, we must have $a_k=0$ for $k\ge6$, leaving
$$\begin{align} a_5&=\big\lfloor{1001\over206}\big\rfloor=4\\ \\ a_4&=\big\lfloor{1001-4\cdot206\over41}\big\rfloor=\big\lfloor{177\over41}\big\rfloor=4\\ \\ a_3&=\big\lfloor{177-4\cdot41\over10}\big\rfloor=\big\lfloor{13\over10}\big\rfloor=1\\ \\ a_2&=\big\lfloor{13-10\cdot1\over3}\big\rfloor=\big\lfloor{3\over3}\big\rfloor=1\\ \\ a_1&=0 \end{align}$$
This gives
$$x=0+2\cdot1+6\cdot1+24\cdot4+120\cdot4=2+6+96+480=584$$
as others have found.
Remark: The restriction $0\le a_k\le k$ makes it clear that not every number $N$ can be written in the form $\big\lfloor {x\over1}\big\rfloor+\big\lfloor {x\over2}\big\rfloor+\big\lfloor {x\over6}\big\rfloor+\big\lfloor {x\over24}\big\rfloor+\cdots$ (with integer $x$). The sequence of numbers that can be so written is A060832 in the OEIS. As it happens, $N=1001$, as we just saw, is such a number, but $N=1000$ is not.
$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\lfloor\frac{{x}}{{n}!}\rfloor=\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\frac{{x}}{{n}!}\:−{f}_{{n}} \right)=\mathrm{1001} \\ $$ $${x}\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\frac{\mathrm{1}}{{n}!}\right)=\left[\mathrm{1001};\mathrm{1011}\right) \\ $$ $$\lfloor{x}\rfloor=\left[\mathrm{342};\mathrm{345}\right],\lfloor\frac{{x}}{\mathrm{2}!}\rfloor=\left\{\mathrm{171},\mathrm{172}\right\} \\ $$ $$\lfloor\frac{{x}}{\mathrm{1}!}\rfloor+\lfloor\frac{{x}}{\mathrm{2}!}\rfloor=\mathrm{1001}−\underset{{n}=\mathrm{3}} {\overset{\mathrm{10}} {\sum}}\lfloor\frac{\mathrm{342}}{{n}!}\rfloor \\ $$ $$\lfloor{x}\rfloor+\lfloor\frac{{x}}{\mathrm{2}!}\rfloor=\mathrm{928}\:\left[{NO}\:{SOLUTION}\right] \\ $$
