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The following conversion equation appears to work for all positive integers. I verified this experimentally. $$ x=\left\lfloor\left\lceil x\cdot \frac{412}{256}\right\rceil\cdot \frac{256}{412}\right\rfloor $$ I'm not sure how to solve or make progress on this floor and ceiling equation for $x$. Could anyone point me in the right direction for tackling this problem? Or perhaps a counterexample for this if one comes to mind?

Result from Wolframalpha.

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    $\begingroup$ @GSmith, I think he already did: See the Wolfram hyperlink. $\endgroup$
    – Simon
    Commented Jul 10 at 5:35
  • $\begingroup$ @Simon so Wolfram Alpha counts? when i posted the precious comment i thought is wasn't the right result or something, as the text for the hyperlink said wrong result. I wasn't too sure whether wolfram alpha counts as a proof before as well, but thanks for clarifying. $\endgroup$
    – GSmith
    Commented Jul 10 at 5:46
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    $\begingroup$ @GSmith, I made many mistakes because I'm not familiar with this system and English. Looking at the Wolframalpha graph, it seems like it would always be the case, but I couldn't be certain, so I asked this question. $\endgroup$
    – user150497
    Commented Jul 10 at 6:00
  • $\begingroup$ Hint: you can prove it directly using inequalities (for example, start by converting the statement "$x=\lfloor y\rfloor$" into the form $x\le y < x+1$) $\endgroup$ Commented Jul 10 at 8:41

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Let's look in general at the equation $$\tag{*} x=\left\lfloor\frac{\left\lceil ax\right\rceil}{a}\right\rfloor$$ for real positive $a$ (in the case of this question, we have $a=\frac{103}{64}$). The right-hand side is always an integer, so the only possible solutions are integers. We can prove that for certain $a$, the equation holds for all integers $x$.

Starting with the definition of $\left\lceil ax\right\rceil$, $$ax\le \left\lceil ax\right\rceil < ax+1$$ Dividing by $a$, $$x\le \frac{\left\lceil ax\right\rceil}{a} < x+\frac{1}{a}$$

If $a\ge 1$, then we have $$x\le \frac{\left\lceil ax\right\rceil}{a} < x+1$$

which is equivalent to $(*)$ by the definition of floor.

Since $\frac{103}{64}>1$, this proves the conjecture. It's worth noting that for $a<1$, the statement doesn't hold for all integer $x$.

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