Solve for $x$:
$x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor=2001$
Given solution:
$\lfloor x\rfloor=\lfloor \sqrt[4]{2001}\rfloor=\{6,-7\}$
If $\lfloor x\rfloor=6\Rightarrow x\lfloor x\lfloor 6x\rfloor\rfloor=2001$
So far so good. Now the next step was too abrupt for me. The author has written
$\color{red}{\lfloor 6x\rfloor=\lfloor \sqrt[3]{(6)^2(2001)}\rfloor=41}$
$\Rightarrow x\lfloor 41x\rfloor=2001$
Now the author has used the same method as above and has written
$\color{red}{\lfloor 41x\rfloor=\lfloor \sqrt{(41)(2001)}\rfloor=286}$
and thus immediately arrives at the answer $x(286)=2001\Rightarrow x=\frac{2001}{286}$
How does $\color{red}{this}$ happen is what I would want to know. I am missing out on a basic result probably.
Though the question has a solution on StackExchange I want to know about the method I have described above which I find quite unique.
As the user @peterwhy pointed out in Comments the logic could have been $x\lfloor x\lfloor 6x\rfloor \rfloor =2001\Rightarrow 6x.6\lfloor x\lfloor 6x\rfloor \rfloor =6^2.2001$. Could this imply $\lfloor 6x\rfloor^3=6^2.2001\Rightarrow \lfloor 6x\rfloor=\lfloor \sqrt[3]{6^2.2001}\rfloor=41$
Best Answer
From the property that $\lfloor a \rfloor \le a$, the LHS of $(1)$ satisfies
$$\begin{align*} x\lfloor x\lfloor x\cdot 6\rfloor\rfloor &\le x\cdot x\lfloor x\cdot 6\rfloor \le x\cdot x\cdot x\cdot 6\\ x^3\cdot 6 &\ge 2001\\ x\cdot 6 &\ge \sqrt[3]{6^2\cdot2001} \approx 41.6 \end{align*}$$
From $(2)$,
$$\begin{align*} 6 \le x &< 7\\ x\cdot 6 &< 42 \end{align*}$$
So $\lfloor x\lfloor x\rfloor\rfloor = \lfloor x\cdot 6\rfloor = 41$.
Similarly, from $x\lfloor x\cdot41\rfloor = x\lfloor x\lfloor x\cdot 6\rfloor\rfloor = 2001$,
$$\begin{align*} x^2 \cdot 41 &\ge x\lfloor x\cdot41\rfloor = 2001\\ x \cdot 41 &\ge \sqrt{41\cdot 2001} \approx 286.4 \end{align*}$$
And from the previous step,
$$\begin{align*} 41.6\ldots \le x \cdot6 &< 42\\ 6.9\ldots \le x &< 7\\ 284.3\ldots \le x\cdot 41 &< 287 \end{align*}$$
So $\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = \lfloor x\cdot 41\rfloor = 286$.
One may say this step works, with $\lfloor x\cdot 6\rfloor$ narrowed down to one possibility, only because the lower and upper bounds of $x\cdot 6$ is near. This, I would agree.
This was also observed by some answers and comments in a similar question: find positive real number x that satisfies $2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$.
Here $\lfloor x\rfloor = \left\lfloor\sqrt[4]{31}\right\rfloor = 2$, then on one side $2x\ge \sqrt[3]{2^2\cdot 31} \approx 4.99$ while $2x < 2\cdot 3 = 6$. The possibilities for the next step is $\lfloor x\lfloor x\rfloor\rfloor = \lfloor 2x\rfloor \in\{4,5\}$:
$$\begin{array}{r|c|c} \lfloor x\lfloor x\rfloor\rfloor = & 4 & 5 \\\hline \text{Bounds from $x\lfloor x\rfloor$:}\\ x\cdot 2 = x\lfloor x\rfloor \in & [4.99\ldots, 5) & [5,6)\\ \overset{/2}{\implies} x \in & \left[2.49\ldots, \frac52\right) & \left[\frac52, \frac62\right)\\ \overset{\cdot \lfloor x\lfloor x\rfloor\rfloor}{\implies} x\lfloor x\lfloor x\rfloor\rfloor \in & [9.97\ldots, 10) & \left[12.5, 15\right)\\\hline \text{Lower bound from the given:}\\ 31 = x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \le x^2\lfloor x\lfloor x\rfloor\rfloor = & 4x^2 & 5x^2\\ x\lfloor x\lfloor x\rfloor\rfloor \ge & \sqrt{4\cdot31}\approx 11.1 & \sqrt{5\cdot31}\approx 12.4\\\hline \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \in & \emptyset & \{12,13,14\} \end{array}$$
The next step even has 3 possibilities:
$$\begin{array}{r|c|c|c} \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = & 12 & 13 & 14\\\hline \text{Bounds from $x\lfloor x\lfloor x\rfloor\rfloor$:}\\ x\cdot 5 = x\lfloor x\lfloor x\rfloor\rfloor \in & [12.5, 13) & [13, 14) & [14, 15)\\ \overset{/5}{\implies} x \in & [2.5, 2.6) & [2.6, 2.8) & [2.8, 3)\\ \overset{\cdot \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor}{\implies} x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \in & [30, 31.2) & [33.8, 36.4) & [39.2, 42)\\\hline \text{From the given:}\\ x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = & 31 & 31 & 31\\ x = \frac{31}{\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor} = & \frac{31}{12} \approx 2.5\\\hline \text{Verifications:}\\ \lfloor x\rfloor = & \lfloor2.5\ldots\rfloor\\ \lfloor x\lfloor x\rfloor\rfloor = & \lfloor5.1\ldots\rfloor\\ \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = & \lfloor12.91\ldots\rfloor \end{array}$$
So the only positive solution $x$ is $\frac{31}{12}$, which has $\lfloor x\lfloor x\rfloor\rfloor = 5 \ne \left\lfloor\sqrt[3]{2^2\cdot 31}\right\rfloor = 4$.