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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.]

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    What have you tried so far? We would like to help but we need to know where you are getting stuck. – 2017-01-02
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    I think you are lucky to get the upvotes without any context at all. – 2017-01-02
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    To ask a question on this site, there is no need to be stuck or to have a "context". – 2017-01-02
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    [Have you thoroughly searched for an answer before asking your question? Sharing your research helps everyone. Tell us what you found and why it didn’t meet your needs. This demonstrates that you’ve taken the time to try to help yourself, it saves us from reiterating obvious answers, and above all, it helps you get a more specific and relevant answer!](http://math.stackexchange.com/help/how-to-ask) – 2017-01-02

1 Answers 1

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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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    I think you need to fix a small typo, If $2n\le\frac{15n}{9}$ then n is zero or negative “not positive” $\Rightarrow n\le0$. – 2017-01-02
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    @HazemOrabi I'm sorry, but I don't think I'm understanding your point. – 2017-01-02
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    @florence: You said it correctly: positive solutions are ruled out. (+1) – 2017-01-02
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    yes, sorry, U R correct. – 2017-01-02
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    $\lfloor x \rfloor > x - 1$; therefore $\left\lfloor \frac{15n}{9}\right\rfloor > \frac{15n}{9}-1$ which removes the need of manually testing $n=-3$ – 2017-01-02