Calculation of $x$ in $x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88$

Question

How can I calculate real values of $x$ in $x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88$, where $\lfloor x\rfloor$ is the floor function?

My attempt:

Let $\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = k\in \mathbb{Z}$. Then

$$ k \leq x\lfloor x\lfloor x\rfloor\rfloor<k+1 $$

and our equation becomes

$$ x\cdot k = 88 \implies x=\frac{88}{k} $$

For $x>0$, simple guessing shows that $3.1<x<3.2$. But how can we account for $x<0$?

Answer 1

If $0\le x<3$, then the expression is $<3^4<88$. So in the positive case, $k=\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\ge 27$ and also $k=\frac{88}x\le \frac{88}3$, i.e. $k\in\{27,28,29\}$. Trying $x=\frac{88}{28}=\frac{22}{7}$ we verify that we have found a solution, whereas $\frac{88}{27}$ and $\frac{88}{29}$ don't work. (Since $x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$ is strictly decreasing (in the domain $x\ge1$) we might not even fully test other values once we have found or otherwise guessed a solution).

To deal with negative $x$, we might substitue $x$ with $-y$ and investigate $$y\lceil y\lceil y \lceil y\rceil\rceil\rceil=88$$ with $y>0$ instead. Once again, for $0<y\le 3$, the left hand side is $\le 81<88$, hence $y>3$, but then the left hand side is $\ge 64 y>192>88$.

Answer 2

There isn't a reason to neglect negative values, but note that if $x=-3-\epsilon$ with $\epsilon$ small ($0\lt \epsilon \lt \frac 1{37}$) you have successively $$\lfloor x \rfloor=-4$$ $$\lfloor -4(-3-\epsilon) \rfloor=12$$ $$\lfloor 12(-3-\epsilon) \rfloor=-37$$$$-37(-3-\epsilon)\gt111$$ And if $x=-3$ the value is $81$, so you can't attain the value $88$ with a negative value of $x$.

Answer 3

Through guess and check

$x=\dfrac{22}{7}$

We know apriori, $x\approx 3.16 \Longrightarrow \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 28$, so setting $k=28$ yields $x=22/7$

You can verify this yourself:
$$ \tfrac{22}{7}\lfloor\tfrac{22}{7}\lfloor \tfrac{22}{7}\lfloor \tfrac{22}{7}\rfloor\rfloor\rfloor =88 $$

Answer 4

Hint

Let $x=\frac p q\in[k,k+1)$ irreductible and positive so $88\times \frac q p\in\mathbb N$ hence $$p\in\{1,2,4,8,11,22,44,88\}$$ and we have

$$k^4\leq x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88\leq (k+1)^4$$ so $k=3$ and then what's the possible values of $x$? Repeat the same reasoning for $x$ negative.

Answer 5

If there is a rational solution for $x$, we know that its numerator must divide $88$ evenly. This leaves us with $1$, $2$, $4$, $8$, $11$, $22$, $44$, and $88$ as possible choices of numerator for $x$.

It turns out that $x=\frac{22}{7}$. You can check all other divisors of $88$ to verify that none of them work as a numerator besides $22$.