$ \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.

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 $584 \leq x < 585$.
