I found a question where we need to find the number of values satisfying this equation with the constraint that $x \in (0,1000)$

$$\left\lfloor\frac{x}{2}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\left\lfloor\frac{x}{5}\right\rfloor=\frac{31x}{30}$$ where $\left\lfloor.\right\rfloor$ represents the floor function of $x$.

Now here its obvious that the right hand side must also give out an integer hence all the multiples of $30$ would work hence giving $30$ solutions between $0$ and $1000$ the result must be in form of $\frac{30k}{31}$ but how do I find such $k$ so as to give the other solution of $x$. Is there any method to solve the equation directly? Any hint would work.

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    $\begingroup$ I would have thought you have shown $x$ must be an integer multiple of $30$. All these work and there are $\lfloor \frac{1000-1}{30} \rfloor = 33$ of them strictly greater than $0$ and less than $1000$ $\endgroup$ – Henry Oct 28 '16 at 8:00
  • $\begingroup$ $x$ is an integer? Not necessarily, but since the LHS so must be the right, correct? $\endgroup$ – Jimmy R. Oct 28 '16 at 8:10
  • $\begingroup$ @JimmyR. There is no such condition mentioned. $\endgroup$ – Harsh Sharma Oct 28 '16 at 8:12
  • $\begingroup$ @Henry How do I prove there is no value of x in the form of $\frac {30k}{31}$ because this would also give the values of integers in the RHS? $\endgroup$ – Harsh Sharma Oct 28 '16 at 8:14

Suppose $x=\frac{30k}{31}$ is not an integer

then there is rounding down and $\displaystyle \left\lfloor\frac{x}{2}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\left\lfloor\frac{x}{5}\right\rfloor \lt \frac{x}{2}+\frac{x}{3}+\frac{x}{5} = \frac{31x}{30}$

so the only solutions are of the form $x=30k$ for integer $k$


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