Solving equation with two floor functions

Question

I'm trying to solve the following question

\begin{equation*} \left\lfloor \frac{\left\lfloor \frac{3\lfloor x\rfloor }{2}\right\rfloor }{9}\right\rfloor =4 \end{equation*}

I got the following inequalities: $ n \leq x \lt n+1 \\ m \leq \frac{3n}{2} \lt m+1 \\ 36 \leq m \lt 45 \\ ⌊⌋= n \\ ⌊\frac{3n}{2}⌋= m$

I don't know how to deal with the floor functions, so I have no idea where to start. If someone could walk me through the process that would be great!

Answer 1

Note that $\lfloor x\rfloor=n \Leftrightarrow n\leqslant x<n+1$. We can simplify the equation and get inequalities. Further, since floor function can only have values in $\mathbb{Z}$, we can discuss each possible values case by case.

Answer 2

Suppose $\lfloor x\rfloor = n$; as you noted, this means that $n\leq x\lt n+1$. Next, suppose that $\lfloor \frac{3\lfloor x\rfloor}{2}\rfloor = m$; as before, this means that $m\leq \frac{3\lfloor x\rfloor}{2}\lt m+1$; from this, we obtain $\lfloor \frac{m}{9}\rfloor=4$. This leads to $4\leq \frac{m}{9}\lt 5$, which of course means that $36\leq m\lt 45$, or $36\leq m\leq 44$.

Now we can back-substitute, so to speak. If $36\leq m\leq 44$, then we have $36\leq \frac{3\lfloor x\rfloor}{2}\lt 45$, which means that $24\leq \lfloor x\rfloor \lt 30$, or $24\leq \lfloor x\rfloor \leq 29$

In other words, $x\in [24,30)$. You can verify this by checking that $x$ taken from the endpoints of this interval satisfy the equation; given that the floor function is nondecreasing, this means that any $x$ within that interval will satisfy it

Answer 3

Just take deep breathes.

$\left\lfloor \frac{\left\lfloor \frac{3\lfloor x\rfloor }{2}\right\rfloor }{9}\right\rfloor =4$

$4\le \frac{\lfloor \frac {3\lfloor x\rfloor}2\rfloor}9 < 5$

$36 \le \lfloor \frac {3\lfloor x\rfloor}2\rfloor < 45$

So we know that $\lfloor \frac {3\lfloor x\rfloor}2\rfloor$ is an integer and it is an integer between $36$ (inclusive) and $45$ exclusive, so

$36 \le \frac {3\lfloor x\rfloor}2 < 45$

$72 \le 3\lfloor x \rfloor < 90$

$24 \le \lfloor x \rfloor < 30$ and we know $\lfloor x \rfloor$ is an integer between $24$ and $30$ so $x$ maybe be an any real number $x \in [24, 30)$.