$$2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\\ (x>0 ,x\in\mathbb R)$$ find $x$ that satisfies the expression above.
My attempt
Since $$(x+1)^4\gt\text{(right side of the expression given)}\geq x^4\\ 7\gt x\geq6$$

And, $\frac{2001}x$ needs to be a integer.
If $x$ is not a rational number, $\frac{2001}x$ is not a integer.
So, $x$ can be thought as $\frac ab(a,b\in\mathbb Z,a|2001)$
But, I failed to go further.

  • 3
    $\begingroup$ Well, there aren't that many integers of the form $2001/x$ for $6 \le x < 7$. The least is $286$. $\endgroup$ Jan 6, 2020 at 18:35
  • 4
    $\begingroup$ I wonder if it helps to know that the prime factorization of $2001$ is $3\cdot 23\cdot 29$ ? $\endgroup$
    – MPW
    Jan 6, 2020 at 18:35

3 Answers 3


By letting $f(x)=x\lfloor x\lfloor x \lfloor x\rfloor\rfloor\rfloor $ we have $$ \lim_{x\to 7^-}f(x) = 7(7(7(7 - 1) - 1) - 1) = 2002 $$ and in a left neighbourhood of $x=7$ our function is a linear function with derivative $(7(7(7 - 1) - 1) - 1)=286.$ Since $f$ is increasing, the problem boils down to solving $$ 286(x-7)+2002 = 2001 $$ which leads to $x=\frac{2001}{286}$.

  • 2
    $\begingroup$ This works out, but - similarly to the other solution - relies on the assumption that $x$ is very close to $7$, which turns out to be correct. (If $x$ were far from $7$, then $f(x)$ would not be linear in the interval $[x,7]$.) Though the observation that $$\lim_{x \to 7^-} f(x) = 2002$$ which is very close to $2001$ does suggest that probably $x$ is very close to $7$. $\endgroup$ Jan 6, 2020 at 18:58

Using what you have so far, suppose $x= 7 - \epsilon$ where $0\leq \epsilon <1.$ Some computing leads me to believe that $\epsilon$ is quite small. Then

$$x\lfloor x \lfloor x\lfloor x \rfloor \rfloor \rfloor = x\lfloor x \lfloor (7-\epsilon)6 \rfloor \rfloor=x \lfloor x\lfloor 42-6\epsilon\rfloor\rfloor.$$

I might have some case work here, but since I think $\epsilon$ is small, I'll start with the case that $\epsilon < 1/6$ to get the above

$$=x \lfloor x(41)\rfloor = x\lfloor (7-x)41\rfloor = x\lfloor 287 - 41\epsilon \rfloor.$$

Perhaps more case work (I fear there might be 41 cases, but maybe I'll get lucky). Assume $\epsilon < 1/41.$ So the equation becomes

$$2001 = x*286$$

And by golly $x = \frac{2001}{286}$ works.


If you know that $x$ is 'just below' $7$ then, keeping your fingers crossed, just go for it!

The OP's setup allows us to write

$\tag 1 \lfloor x \rfloor = 6$

and you can begin by setting $x$ to $6$ on your 'slider bar'.

Now keep pushing $x \lt 7$ to the right until you can write ($\, 7 \times 6 - 1 = 41\,$)

$\tag 2 \lfloor x \times 6 \rfloor = 41$

Now keep pushing $x \lt 7$ to the right until you can write ($\, 7 \times 41 - 1 = 286\,$)

$\tag 3 \lfloor x \times 41 \rfloor = 286$

You are now left with (the rhs of the OP equation after working from the inside to the outside),

$\tag 4 x \times 286 \lt 7 * 286 = 2002$

Of course if you are saving $x = \frac{286}{41}$ from $\text{(3})$ you can push it further to the right and write

$\tag 5 x \times 286 = 2001$

so that the answer is given by

$\tag 6 x = \frac{2001}{286}$

Extra Credit: Determine if the following two equations have solutions for $x \gt 0$:

$\quad 1996=x⌊x⌊x⌊x⌋⌋⌋$
$\quad 1995=x⌊x⌊x⌊x⌋⌋⌋$

Note that we can generate similar problems.

For example, if we started by saying that $x$ is 'just below' $6$ we can crank out another corresponding 'max integer' $n$ such that

$\quad n = x⌊x⌊x⌊x⌋⌋⌋$

We would find $n$ and then ask the student to solve

$\quad 1037 = x⌊x⌊x⌊x⌋⌋⌋$

The procedure/algorithm being defined can actually be proven to produce well-defined outcomes - there is no reason to plug the 'found' $x$ back into the equation to see that 'it works'.


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