<s>[INST] (floor function) sum of x: $\left\lfloor{\frac{x}{5}}\right\rfloor - \left\lfloor{\frac{x}{9}}\right\rfloor = \frac{x}{15}$ Find the sum of integers $x$ such that

$$\left\lfloor{\frac{x}{5}}\right\rfloor - \left\lfloor{\frac{x}{9}}\right\rfloor = \frac{x}{15}$$

I don't have much experience in solving problems with the floor function involved. I have rewritten the equation as

$$\frac{x}{45} + \left\{\frac{5x}{45}\right\} = \left\{\frac{9x}{45}\right\}$$

but don't know where to go from there.
[EDIT: Thanks, I understand how to solve it now.]
 [/INST] First, since the LHS is an integer, we know $x$ is a multiple of $15$. Thus, let $x = 15n$. Then our equation reduces to
$$n = 3n - \left\lfloor \frac{15n}{9}\right\rfloor \implies$$
$$2n = \left\lfloor \frac{15n}{9}\right\rfloor \leq \frac{15n}{9}$$
thus ruling out any positive solutions. Now note that 
$$2n = \left\lfloor \frac{15n}{9}\right\rfloor \geq \frac{15n}{9}-1 \implies$$
$$n \geq -3$$
So any possible solutions for $n$ satisfy $-3 \leq n \leq 0$. Checking manually, we see that the only solutions are $n = 0, -1, -2$, i.e. $x = 0, -15, -30$.
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