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

Question

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.

Answer 1

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 $584 \leq x < 585$.

Answer 2

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.

Answer 3

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.