# Solve $\frac{1}{\left\lfloor{x}\right\rfloor}+\frac{1}{\left\lfloor{2x}\right\rfloor}=\{x\}+\frac13$ [duplicate]

Problem Statement:-

Solve: $$\dfrac{1}{\left\lfloor{x}\right\rfloor}+\dfrac{1}{\left\lfloor{2x}\right\rfloor}=\{x\}+\dfrac{1}{3}$$ where $$\left\lfloor{x}\right\rfloor$$ denotes the integral part of $$x$$ and $$\{x\}=x-\left\lfloor{x}\right\rfloor$$

The floor function just eats me up whole whenever I encounter it. I have tried some things to solve it but am just not able to come up with anything useful at all. Here are the things I have tried.

First of all lets define the domain of $$x$$ which is $$x\not\in[0,1)$$, because if so happens then $$\left\lfloor{x}\right\rfloor=0$$ which should not happen for $$\dfrac{1}{\left\lfloor{x}\right\rfloor}$$ and $$\dfrac{1}{\left\lfloor{2x}\right\rfloor}$$ to be defined.

After this I was not able to come up with a good line of thought for attempting the problem so, please guide me as to what should be my line of thought for attempting problems related to the integral part of $$x$$.

## merged by quid♦Mar 11 at 14:37

This question was merged with Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$ because it is an exact duplicate of that question.