If $x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor=2018$, find $x$.
My working: if $x$ is positive then by estimation it must be in $(6,7)$ and for this interval I : $x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor<2003$. So no solution for positive $x$.
For negative $x$, I got $x \in (-7,-6)$ and using this I got $x=\frac{2018}{k}, k$ can be $-336,-335,\cdots,-289,$
Now using excel sheet I got $x=-\frac{2018}{305}$.
Best Answer
Let $$f(x)=x\Bigl\lfloor x\bigl\lfloor x\lfloor x\rfloor\bigr\rfloor\Bigr\rfloor. $$ Note that $f$ is monotonic on $[1,\infty)$ and monotonic on $(-\infty,0]$, whereas there is clearly no solution with $0<x<1$ (where $f(x)=0$). Hence there is possibly one positive and/or one negative solution.
We have $f(x)\approx x^4$ so we solve $x^4=2018$ instead, which has two real solutions $$x=\pm\sqrt[4]{2018}=\pm 6.702\ldots$$ so that we suspect $$\lfloor x\rfloor = 6\qquad\text{or}\qquad \lfloor x\rfloor=-7$$ Then $f(x)\approx 6x^3$ or $\approx -7x^3$ depending on the sign of $x$, so we solve $6x^3=2018$ and $-7x^3=2018$ instead. Each gives us one real solution $$ x=\sqrt[3]{\frac{2018}6}\qquad\text{resp.}\qquad x =-\sqrt[3]{\frac{2018}7}$$ so that $$6x=41.726\ldots\qquad\text{resp.}\qquad -7x=46.242\ldots$$ so that we suspect $$\lfloor 6x\rfloor =41\qquad\text{resp.}\qquad \lfloor -7x\rfloor =46.$$ Then $f(x)\approx 41x^2$ with $x>0$ or $\approx 46x^2$ with $x<0$. We solve these approximations and find $$ x=\sqrt{\frac{2018}{41}}\qquad\text{resp.}\qquad x=-\sqrt{\frac{2018}{46}}$$ so that $$ 41x=287.642\ldots\qquad\text{resp.}\qquad 46x=-304.676\ldots$$ and we suspect $$\lfloor 41x\rfloor = 287\qquad\text{resp.}\qquad\lfloor 46x\rfloor = -305. $$ Now for such $x$ we have exactly $f(x)=287x$ resp. $f(x)=-305x$, so we need $x=\frac{2018}{287}$ or $$\fbox{$\displaystyle{x=-\frac{2018}{305}}$}.$$ The latter turns out to be a solution and hence is the only negative solution. However, the former does not work out: $f(\frac{2018}{287})>2018$. Let's revisit our approximations. As the floor function decreases values, but by less than $1$, we have $f(x)=(x-\delta)^4$ for some $0\le \delta<1$ (at least for $x\ge 1$). So $x-\delta=6.702\ldots$ and either our guess $\lfloor x\rfloor =6$ was correct, or we have $\lfloor x\rfloor =7$ - but then $f(x)\ge f(7)>2018$. So we do have $f(x)=x\bigl \lfloor x\lfloor 6x\rfloor \bigr\rfloor$ as used above. Then in the range of interest, we do have $f(x)=6(x-\delta)^3$ with $0\le \delta<\frac 16$. So $6x-6\delta=41.726\ldots$ and either $\lfloor6x\rfloor = 41$ or $\lfloor 6x\rfloor =42$ - but the latter would mean $x\ge 7$, which we already excluded. Hence our guess $\lfloor6x\rfloor = 41$ must have been correct. Now $f(x)=x\lfloor41x\rfloor$ and we have $f(x)=41(x-\delta)^2$ for some $0\le\delta<\frac1{41}$. Then $41x-41\delta=287.642\ldots$ and so either $\lfloor 41x\rfloor =287$ or $\lfloor41\rfloor=288$. Then $x=\frac{2018}{287}$ (excluded above) or $x=\frac{2018}{288}$ - but once again the latter is $>7$. So there indeed is no positive solution.