find positive real number x that satisfies $2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$ $$2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\\
(x>0 ,x\in\mathbb R)$$
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.
 A: 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 $x=\frac{2001}{286}$.
A: 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.  
A: 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'.
