x⌊x⌊x⌊x⌋⌋⌋ = 2020

Question

Solve for $x$: $$ x \left\lfloor x \left\lfloor x \left\lfloor x \right\rfloor \right\rfloor \right\rfloor = 2020. $$

The floor function $\left\lfloor t \right\rfloor$ has the usual “greatest integer $\leq t$” definition.

Attribution pending From Michael Penn’s Solving a crazy iterated floor equation video. He proposes four slight variations of the problem at the very end.

Answer 1

Answer:

$x=-\frac{2020}{305}=-\frac{404}{61}$

Explanation:

Firstly, let's notice that $x$ multiplied by an integer gives $2020$, so we have $x=\frac{2020}{\alpha}$ for some integer $\alpha$. Since $6^4=1296<2020<2401=7^4$, the value of $|x|$ must be between $6$ and $7$ (that's because the function is increasing for positive $x$ and decreasing for negative $x$). So, $|\alpha|$ must be between $288$ and $337$. Now we just can bruteforce all values using simple Python code: Try it online! and find $\alpha=-305$, the only suitable value.

Answer 2

Observations to give lower and upper bounds:

1. $\lfloor x\rfloor\leq x$, so we must have $2020=x \left\lfloor x \left\lfloor x \left\lfloor x \right\rfloor \right\rfloor \right\rfloor\leq x^4$, therefore $x\geq\sqrt[4]{2020}=6.704$

2. If $x\geq7$, then $x\lfloor x\rfloor\geq49$ and so on until $2020=x \left\lfloor x \left\lfloor x \left\lfloor x \right\rfloor \right\rfloor \right\rfloor\geq 7^4=2401$. Contradiction.

So we know for sure

$x$ is six point something and $\lfloor x\rfloor=6$. Also $6.704\leq x<7$ means $40.224\leq6x<42$, so $\lfloor x\lfloor x\rfloor\rfloor$ must be either $40$ or $41$.

Now the whole thing becomes

$2020=x \left\lfloor x (40\text{ or }41) \right\rfloor$. The thing inside this final floor sign is at least $40\times6.704=268.16$ and at most $41\times7=287$. Which means $x$ must be at least $2020\div287=7.038$.

Contradiction ... and now I realise my implicit assumption that

$x\geq0$.

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Going back to those two observations at the beginning,

with the knowledge that $x$ is negative, we have $$x\geq-6\Rightarrow\lfloor x\rfloor\geq-6\Rightarrow x\lfloor x\rfloor\leq36\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\geq-216\Rightarrow x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\leq 1296,$$ $$x\leq-7\Rightarrow\lfloor x\rfloor\leq-7\Rightarrow x\lfloor x\rfloor\geq49\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\leq-343\Rightarrow x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\geq 2401,$$ $$\lfloor x\rfloor\leq x\Rightarrow x\lfloor x\rfloor\geq x^2\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\leq x\lfloor x^2\rfloor\leq x^3\Rightarrow 2020=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\geq x^4,$$ so $-6.704\leq x<-6$ and $\lfloor x\rfloor=-7$.

That means $42<x\lfloor x\rfloor\leq46.928$ and $43\leq\lfloor x\lfloor x\rfloor\rfloor\leq46$.

That means $-308.38\leq x\lfloor x\lfloor x\rfloor\rfloor<-258$.

So we seek a number which,

when multiplied by an integer between $258$ and $308$, gives $2020$. Dividing $2020$ by $6$ and $7$ gives that this integer must be between $289$ and $336$. Going the other way, the bound of $308$ means $x\geq-\frac{2020}{308}=-6.558$. Since this bound came from taking the fourth root, we expect $x$ should be close to it.

So we try just a few nearby values of the integer:

$x=-\frac{2020}{308}=-6.558\Rightarrow x\lfloor x\rfloor=7\times6.558=45.909\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-45\times6.558=295.13$, too small.

$x=-\frac{2020}{307}=-6.580\Rightarrow x\lfloor x\rfloor=7\times6.580=46.059\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.580=302.67$, too small but much closer!

$x=-\frac{2020}{306}=-6.601\Rightarrow x\lfloor x\rfloor=7\times6.601=46.209\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.601=303.66$, too small.

$x=-\frac{2020}{305}=-6.623\Rightarrow x\lfloor x\rfloor=7\times6.623=46.361\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.623=304.66$, exactly right!

And we have the solution,

$x=-\frac{2020}{305}=-6.623\dots$

Answer 3

Let us denote $$\aleph(x)=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\,.$$Since the fourth root of $2020$, $\sqrt[4\,]{2020}$, is located between 6 and 7, the solution $x^\star$ is either $x^\star=6+ε$ or $x^\star=-7+ε$, with $ε \in (0,1)$. The puzzle then becomes solving either

\begin{align}\aleph(6+ε) &=(6+ε)⌊(6+ε)⌊(6+ε)⌊6+ε⌋⌋⌋\\ &= (6+ε)⌊(6+ε)⌊36+6ε⌋⌋ \\ &= (6+ε)⌊(6+ε)(36+⌊6ε⌋)⌋ \\&= 2020\end{align}

where there are 6 possible integer values for $⌊6ε⌋$, with only $⌊6ε⌋=5$ being possible, since $\aleph(6+\frac{5}{6})<2020$, turning the equation into

$$(6+ε)⌊41(6+ε)⌋ = (6+ε)(246+⌊41ε⌋) = 2020$$

where again only $⌊41ε⌋=40$ being possible, as $\aleph(6+\frac{40}{41 })<2020$, ending up with

$$1716+286ε = 2020$$

which has no solution in $(\frac{40}{41},1)$.

Hence, moving to the alternative case \begin{align}\aleph(-7+ε) &=(-7+ε)⌊(-7+ε)⌊(-7+ε)⌊-7+ε⌋⌋⌋\\ &= (-7+ε)⌊(-7+ε)(49+⌊-7ε⌋)⌋ \\&= 2020\end{align}

shows that only $⌊-7ε⌋=-3$ is possible, since

$$\aleph(-7+\textstyle{\frac{2}{7}})>2020>\aleph(-7+\textstyle{\frac{3}{7}})$$

leading to

$$(-7+ε)⌊46(-7+ε))⌋ = (-7+ε) (-322+⌊46ε⌋)=2020$$

with only $⌊46ε⌋=17$ possible, as

$$\aleph(-7+\textstyle{\frac{17}{46}})>2020>\aleph(-7+\textstyle{\frac{18}{46}})$$

hence

$$2135-305ε=2020$$

and

$$ε=\frac{115}{305}$$

meaning

$$x^\star=-7+\frac{115}{305} = -\frac{2020}{305}$$

Answer 4

A solution which doesn't require brute-forcing with a computer:

(assuming x < 0, since x > 0 turns out to have no solutions)

-7 < x < -6, so ⌊x⌋ = -7

Now we have

x⌊x⌊-7x⌋⌋ = 2020

let x = -7 + p/7, p∈(0,7) (not necessarily an integer)

We can check (by plugging p=2 and p=3 into the original equation) that 2 < p < 3 thus ⌊-7x⌋ = -7(-7+3/7) = 46. Now we have

x⌊46x⌋ = 2020

let x = -7 + q/46, q∈(0,46)

We know that 2 < 7q/46 < 3, meaning 13 < q < 20. Trying a few values, we see 17 < q < 18, which gives us -305x = 2020

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(this is not my solution. I rephrased it from the comment here)