Finding $x$. The summation of the floor of the equation. I would appreciate if somebody could help me with the following problem

Q:Finding $x$. The summation of the floor of the equation.

$$\sum_{i=1}^{2013}\left\lfloor\frac{x}{i!}\right\rfloor=1001$$
Factorial increases very quickly, so because of the floor, most of the terms are zero. If we solve the equation without floors and only five first terms we get:

$$x+x/2+x/6+x/24+x/120= \frac{103x}{60}= 1001 \Leftrightarrow x \approx 583 $$

Now plugging this back into the original equation we see only five first terms are non-zero and sum gives 999. Increasing $x$ by one to $584$ gives the correct answer.
We have the series of the exponential function:

$$\sum_{n=0}^\infty\left\lfloor\frac{1}{n!}\right\rfloor=e$$

In our case, take the X outside, as a constant, and you get that:

$$x * \sum_{i=1}^\infty\left\lfloor\frac{1}{i!}\right\rfloor=e-1$$

Note, the -1 appears when you take the first element, i=0 out of the sum. As i goes >> 9, each element of the sum goes very fast to 0, so going to infinity or 2013, makes no difference. Thus, the result goes:

$$x = \frac{1001}{e-1}$$
Which is approximately 582.55