Can you solve it?

Let $\displaystyle f(x) = \sum_{n=1}^{10} \left \lfloor \dfrac x{n!} \right \rfloor$ . We note that $\displaystyle \sum_{n=1}^{10} \frac{1}{n!} \approx 1.7$ , $x \implies \approx \dfrac{1001}{1.7} \approx 588$ . Since $588 < 6! < 7!...<10!$ , for $f(x) = 1001$ in effect $\displaystyle f(x) = \sum_{n=1}^{\color{#D61F06}{5}} \left \lfloor \dfrac x{n!} \right \rfloor$ . So, we have:

$\begin{aligned} f(588) & = \left \lfloor \dfrac {588}{1!} \right \rfloor + \left \lfloor \dfrac {588}{2!} \right \rfloor + \left \lfloor \dfrac {588}{3!} \right \rfloor + \left \lfloor \dfrac {588}{4!} \right \rfloor + \left \lfloor \dfrac {588}{5!} \right \rfloor \\ & = \left \lfloor \dfrac {588}{1} \right \rfloor + \left \lfloor \dfrac {588}{2} \right \rfloor + \left \lfloor \dfrac {588}{6} \right \rfloor + \left \lfloor \dfrac {588}{24} \right \rfloor + \left \lfloor \dfrac {588}{120} \right \rfloor \\ & = 588 + 294 + 98 + 24 + 4 \\ & = 1008 \end{aligned}$

We note that for every increment of $x$ ,

• $\Delta x = 1 \implies \Delta f(x) = 1$
• $\Delta x = 2 \implies \Delta f(x) = 3$
• $\Delta x = 6 \implies \Delta f(x) = 10$
• Since $6|588$ or $588$ is a multiple of $6$ $\implies f(588-6) = 1008 - 10$ or $f(582) = 998$
• For $f(582+\Delta x) = 988+3\implies \Delta f(x) = 3 \implies \Delta x = 2$
• Therefore $x = 582+2 = \boxed{584}$

It is not hard to see that

$x<6!$ .

Therefore all terms after $[x/5!]$ cancel out to be 0. Since all after that have integral part 0.

Now we see that $x \in(500,600)$ .

Thus $[x/5!]$ = 4.

Now with remaining

$[x/1!] + [x/2!]+ [x/3!]+ [x/4!]=997$ .

Now further checking gives that

$x \in(576,600)$ .

Now we check for 582( can you guess the logic?)

We get $x \in( 582,600)$ .

We find \( [x/24] = 24 , [x/6]=97).

Thus \( [x/1!] + [x/2!]= 876 \).

Now we check for 584 ( using the same logic) and we get our answer.

By simple trial and error; $x=5!+n$ such that $5!<5!+n<6!$

Substituting this into the original equation becomes:

$(5!+n)+(60+\left\lfloor \frac { n }{ 2! } \right\rfloor )+(20+\left\lfloor \frac { n }{ 3! } \right\rfloor )+(5+\left\lfloor \frac { n }{ 4! } \right\rfloor )+(1+\left\lfloor \frac { n }{ 5! } \right\rfloor )=1001$

Which becomes:

$n+\left\lfloor \frac { n }{ 2! } \right\rfloor +\left\lfloor \frac { n }{ 3! } \right\rfloor +\left\lfloor \frac { n }{ 4! } \right\rfloor +\left\lfloor \frac { n }{ 5! } \right\rfloor =795$

And we can just play around with this equation to get the answer.