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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 $x=-\frac{2018}{305}$.

Theo Bendit
  • 50,900
Makar
  • 2,267

3 Answers3

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$$x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor=2018$$

Let $f(x) = x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor$

Let $\xi = -\dfrac{6862735}{1037232} = -6 \dfrac{639343}{1037232}$

Then $f(\xi) - 2018 = -\dfrac{1}{1037232} = -\dfrac{1}{2^4 3^3 7^4}$

Let $x = \xi + \delta$. I want to solve $f(x)-2018 = 0$ for $\delta$.

I will assume that $\delta$ is very small. What I mean by that will make itself clear as I progress.

$\lfloor x \rfloor = -7$

$x \lfloor x \rfloor = -7\xi - 7\delta = 46 \dfrac{46639}{148176} - 7\delta$

$\lfloor x \lfloor x \rfloor \rfloor = 46$

$x \lfloor x \lfloor x \rfloor \rfloor = 46\xi + 46 \delta = -304 \dfrac{183641}{518616} + 46 \delta$

$\lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor = -305$

$x \lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor = -305\xi -305\delta = 2017 \dfrac{1037231}{1037232} - 305 \delta $

\begin{align} f(x) &= 0 \\ \left(2017 \dfrac{1037231}{1037232} - 305 \delta \right) - 2018 &= 0 \\ 305 \delta &= -1/1037232 \\ \delta &= -\dfrac{1}{316355760} \end{align}

$\color{red}{x = \xi + \delta = -\dfrac{2018}{305} = -6 \dfrac{188}{305}}$

Steven Alexis Gregory
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1

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.

Hagen von Eitzen
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$$\mathrm{2018}={x}\lfloor{x}\lfloor{x}\lfloor{x}\rfloor\rfloor\rfloor \\ $$ $$\lfloor{x}\rfloor=\lfloor\pm\sqrt[{\mathrm{4}}]{\mathrm{2018}}\rfloor=\left\{-\mathrm{7},\mathrm{6}\right\} \\ $$ $${if};\:\lfloor{x}\rfloor=-\mathrm{7} \\ $$ $$\mathrm{2018}={x}\lfloor{x}\lfloor-\mathrm{7}{x}\rfloor\rfloor \\ $$ $$\lfloor-\mathrm{7}{x}\rfloor=\lfloor\sqrt[{\mathrm{3}}]{\left(-\mathrm{7}\right)^{\mathrm{2}} \left(\mathrm{2018}\right)}\rfloor=\mathrm{46} \\ $$ $$\mathrm{2018}={x}\lfloor\mathrm{46}{x}\rfloor \\ $$ $$\lfloor\mathrm{46}{x}\rfloor=\lfloor-\sqrt{\left(\mathrm{46}\right)\left(\mathrm{2018}\right)}\rfloor=-\mathrm{304} \\ $$ $$\lfloor{x}\rfloor=\lfloor\frac{\mathrm{2018}}{-\mathrm{304}}\rfloor=-\mathrm{7} \\ $$ $${if};\:\lfloor{x}\rfloor=\mathrm{6} \\ $$ $$\mathrm{2018}={x}\lfloor{x}\lfloor\mathrm{6}{x}\rfloor\rfloor \\ $$ $$\lfloor\mathrm{6}{x}\rfloor=\lfloor\sqrt[{\mathrm{3}}]{\left(\mathrm{6}\right)^{\mathrm{2}} \left(\mathrm{2018}\right)}\rfloor=\mathrm{41} \\ $$ $$\mathrm{2018}={x}\lfloor\mathrm{41}{x}\rfloor \\ $$ $$\lfloor\mathrm{41}{x}\rfloor=\lfloor\sqrt{\left(\mathrm{41}\right)\left(\mathrm{2018}\right)}\rfloor=\mathrm{287} \\ $$ $$\lfloor{x}\rfloor=\lfloor\frac{\mathrm{2018}}{\mathrm{287}}\rfloor\neq\mathrm{6}\Rightarrow{x}\notin{Z}^{+} \\ $$ $${x}=-\:\frac{\mathrm{2018}}{\mathrm{304}} \\ $$

yoga mulyadi
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