$ \biggl\lfloor{\frac{x}{1!}\biggr\rfloor} + \biggl\lfloor{\frac{x}{2!}\biggr\rfloor} + \cdots \biggl\lfloor{\frac{x}{10!}\biggr\rfloor}=1001$ 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.

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.
