<s>[INST] Find range of $x$ satisfying $\left \lfloor \frac{3}{x} \right \rfloor+\left \lfloor \frac{4}{x} \right \rfloor=5$ Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function
My try:
As far as domain of LHS is concerned we have $x \ne 0$ and since RHS is positive, we have $x \gt 0$
Now since LHS is sum of two positive integers, let us suppose:
$$\left \lfloor \frac{3}{x} \right \rfloor=m$$ and
$$\left \lfloor \frac{4}{x} \right \rfloor=5-m$$
Thus we have:
$$ m \le \frac{3}{x} \lt m+1$$
$$5-m \le \frac{4}{x} \lt 6-m$$
Adding both we get:
$$5 \le \frac{7}{x} \lt 7$$
$\implies$
$$1 \lt x \le \frac{7}{5}$$
Hence $$x \in (1, 1.4]$$
But answer in book is given as $$x \in (1,\frac{4}{3})$$
What went wrong? 
 [/INST] Since $\frac{4}{x}>\frac{3}{x}$, we have three cases:


*

*$\left [ \frac{3}{x} \right ]=0$ and $\left [ \frac{4}{x} \right ]=5.$ Easy to show that it's impossible.

*$\left [ \frac{3}{x} \right ]=1$ and $\left [ \frac{4}{x} \right ]=4,$ which is impossible again and

*$\left [ \frac{3}{x} \right ]=2$ and $\left [ \frac{4}{x} \right ]=3,$ which gives the answer: $\left(1,\frac{4}{3}\right]$.
Can you end it now?
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