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I'm using the following equation: $$x=5\left\lceil\frac{mx+b}{5}\right\rceil$$ where $m$ and $b$ are unknown reals, though $0<m<1$ is known. I need to find the minimum $x$ such that this is true, in terms of $m$ and $b$. With what equation could I do this? I tried using $x=\left\lfloor\frac{b+5}{1-m}\right\rfloor$, but that didn't work.

Example: $m=0.3$, $b=1000$. The minimum value of $x$ for which this equation is true is $1430$ (I figured this out by graphing). But how would I know that?

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Write $x=5k$ so that $k=\lceil km+\dfrac{b}{5} \rceil$.

That is, $k-1<km+\dfrac{b}{5} \leq k$; now subtract $k+b/5$.

$-1-\dfrac{b}{5}<k(m-1)\leq-\dfrac{b}{5}$, which is: $\dfrac{b+5}{5(1-m)}>k\geq\dfrac{b}{5(1-m)}$.

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    $\begingroup$ I think your bounds may have the wrong sign. If you plug in b and m from my example, you get negative lower bound for k, which is false. Recall that (m-1) will be negative. $\endgroup$
    – Hello Goodbye
    May 6, 2020 at 16:00
  • $\begingroup$ Yes, that was a typo, which is corrected now. $\endgroup$
    – Aravind
    May 6, 2020 at 17:27
  • $\begingroup$ Cool, thanks for your answer! It would feel a bit more complete to re-sub k=x/5, but this is exactly what I was looking for. $\endgroup$
    – Hello Goodbye
    May 6, 2020 at 17:46

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