Finding the interval of $x$ such that $\left\lfloor \frac{4}{x}\right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor = 7$

Question

This is a problem that I came across in my book.

Find the interval in which $x$ must lie for : $$\left\lfloor \frac{4}{x}\right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor = 7$$

To find the interval's upper and lower limit I found the value for which $\frac{4}{x} + \left\lfloor \frac{2}{x} \right\rfloor$ equals $7$ and that for which the same expression equals $8$. Finding no value for the latter, I jumped on to $9$.

This method holds true perhaps because there is no such value for which the expression lies in interval $\big[8,9\big)$ but for sake of completeness I wish to know how to determine whether a sum of two floor functions will have a value in any particular interval.

Any other insights into the method of solving problems of this sort are more than welcome.

The interval for $x$ for which the expression holds true comes out to be $\big(\frac{2}{3},\frac{4}{5}\big]$.

Answer 1

Let $\;\frac{2}{x} =n+f\;$ where $\;n\;$ is a integer and $\;0\le f<1\;.$

If $\;0\leqslant f<\frac12\;$ then $\;\left\lfloor\frac2x\right\rfloor=n\;$ and $\;\left\lfloor\frac4x\right\rfloor=2n\;.$
If $\;\frac12\leqslant f<1\;$ then $\;\left\lfloor\frac2x\right\rfloor=n\;$ and $\;\left\lfloor\frac4x\right\rfloor=2n+1\;.$

For $\;0\le f<\frac12\;$ there is no solution but for $\;\frac12\leqslant f<1\;$ we have $\;n=2\;$ as a solution.

Therefore $\;\frac2x=2+f\;$ with $\;\frac12\leqslant f<1\;$ satisfies the equation.

Hence,

$\frac52\leqslant\frac2x<3\;\;,$

$\frac13<\frac x2\leqslant\frac25\;\;,$

$\frac23<x\leqslant\frac45\;.$

Consequently, the interval for $x$ for which the expression holds true comes out to be $\left(\frac23,\frac45\right].$

Answer 2

Put $x=\frac 2k$ then you are left with $$\lfloor 2k\rfloor +\lfloor k\rfloor =7$$

Since the expression $\lfloor 2k\rfloor +\lfloor k\rfloor =7$ is increasing in $k$ and constant on $[n, n+0.5)$ and $[n+0.5,n+1)$ for all integers $n$. From this it's easy to see that $k=2.5$ and the entire interval $[2.5,3)$ is the solution.

Putting this back we get $(\frac 23,\frac2{2.5}]$.

Answer 3

$$\left\lfloor \frac 2x \right\rfloor + \left\lfloor \frac 4x \right\rfloor = 7$$

Let $\dfrac 2x = n + \epsilon$ where $n \in \Bbb Z$ and $0 \le \epsilon < 1$

Then $\dfrac 4x = 2n + 2\epsilon$ where $2n \in \Bbb Z$ and $0 \le 2\epsilon < 2$

So $\left \lfloor \dfrac 4x \right \rfloor$ equals either $2n$ or $2n+1$.

Since $n + 2n = 7$ has no integer solution, then $\dfrac 12 \le \epsilon < 1$ and $n + (2n+1) = 7$. Hence $n = 2$ and

$$\dfrac 2x \in 2 + \left[\dfrac 12, 1 \right)$$

$$\dfrac 2x \in \left[\dfrac 52, 3 \right)$$

$$ \dfrac x2 \in \left(\dfrac 13, \dfrac 25 \right]$$

$$ x \in \left(\dfrac 23, \dfrac 45 \right]$$