# find positive real number x that satisfies $2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$

$$2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\\ (x>0 ,x\in\mathbb R)$$ find $$x$$ that satisfies the expression above.
My attempt
Since $$(x+1)^4\gt\text{(right side of the expression given)}\geq x^4\\ 7\gt x\geq6$$

And, $$\frac{2001}x$$ needs to be a integer.
If $$x$$ is not a rational number, $$\frac{2001}x$$ is not a integer.
So, $$x$$ can be thought as $$\frac ab(a,b\in\mathbb Z,a|2001)$$
But, I failed to go further.

• Well, there aren't that many integers of the form $2001/x$ for $6 \le x < 7$. The least is $286$. Jan 6, 2020 at 18:35
• I wonder if it helps to know that the prime factorization of $2001$ is $3\cdot 23\cdot 29$ ?
– MPW
Jan 6, 2020 at 18:35

By letting $$f(x)=x\lfloor x\lfloor x \lfloor x\rfloor\rfloor\rfloor$$ we have $$\lim_{x\to 7^-}f(x) = 7(7(7(7 - 1) - 1) - 1) = 2002$$ and in a left neighbourhood of $$x=7$$ our function is a linear function with derivative $$(7(7(7 - 1) - 1) - 1)=286.$$ Since $$f$$ is increasing, the problem boils down to solving $$286(x-7)+2002 = 2001$$ which leads to $$x=\frac{2001}{286}$$.

• This works out, but - similarly to the other solution - relies on the assumption that $x$ is very close to $7$, which turns out to be correct. (If $x$ were far from $7$, then $f(x)$ would not be linear in the interval $[x,7]$.) Though the observation that $$\lim_{x \to 7^-} f(x) = 2002$$ which is very close to $2001$ does suggest that probably $x$ is very close to $7$. Jan 6, 2020 at 18:58

Using what you have so far, suppose $$x= 7 - \epsilon$$ where $$0\leq \epsilon <1.$$ Some computing leads me to believe that $$\epsilon$$ is quite small. Then

$$x\lfloor x \lfloor x\lfloor x \rfloor \rfloor \rfloor = x\lfloor x \lfloor (7-\epsilon)6 \rfloor \rfloor=x \lfloor x\lfloor 42-6\epsilon\rfloor\rfloor.$$

I might have some case work here, but since I think $$\epsilon$$ is small, I'll start with the case that $$\epsilon < 1/6$$ to get the above

$$=x \lfloor x(41)\rfloor = x\lfloor (7-x)41\rfloor = x\lfloor 287 - 41\epsilon \rfloor.$$

Perhaps more case work (I fear there might be 41 cases, but maybe I'll get lucky). Assume $$\epsilon < 1/41.$$ So the equation becomes

$$2001 = x*286$$

And by golly $$x = \frac{2001}{286}$$ works.

If you know that $$x$$ is 'just below' $$7$$ then, keeping your fingers crossed, just go for it!

The OP's setup allows us to write

$$\tag 1 \lfloor x \rfloor = 6$$

and you can begin by setting $$x$$ to $$6$$ on your 'slider bar'.

Now keep pushing $$x \lt 7$$ to the right until you can write ($$\, 7 \times 6 - 1 = 41\,$$)

$$\tag 2 \lfloor x \times 6 \rfloor = 41$$

Now keep pushing $$x \lt 7$$ to the right until you can write ($$\, 7 \times 41 - 1 = 286\,$$)

$$\tag 3 \lfloor x \times 41 \rfloor = 286$$

You are now left with (the rhs of the OP equation after working from the inside to the outside),

$$\tag 4 x \times 286 \lt 7 * 286 = 2002$$

Of course if you are saving $$x = \frac{286}{41}$$ from $$\text{(3})$$ you can push it further to the right and write

$$\tag 5 x \times 286 = 2001$$

so that the answer is given by

$$\tag 6 x = \frac{2001}{286}$$

Extra Credit: Determine if the following two equations have solutions for $$x \gt 0$$:

$$\quad 1996=x⌊x⌊x⌊x⌋⌋⌋$$
$$\quad 1995=x⌊x⌊x⌊x⌋⌋⌋$$

Note that we can generate similar problems.

For example, if we started by saying that $$x$$ is 'just below' $$6$$ we can crank out another corresponding 'max integer' $$n$$ such that

$$\quad n = x⌊x⌊x⌊x⌋⌋⌋$$

We would find $$n$$ and then ask the student to solve

$$\quad 1037 = x⌊x⌊x⌊x⌋⌋⌋$$

The procedure/algorithm being defined can actually be proven to produce well-defined outcomes - there is no reason to plug the 'found' $$x$$ back into the equation to see that 'it works'.