Assume that $A$, $B$, $C$ are positive real numbers and that $I$ is a positive integer. How could I isolate $A$ in the inequality $\left \lfloor{AB/C}\right \rfloor \geq I$ ? The best I could do:

\begin{equation} AB/C -1 \geq I \Rightarrow \left \lfloor{AB/C}\right \rfloor \geq I \end{equation}

Thus, $A\geq C(1+I)/B$ is a sufficient condition for $\left \lfloor{AB/C}\right \rfloor \geq I$.

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    $\begingroup$ You could simplify your problem by setting $D:=B/C$ and just ask a question about $\left \lfloor{AD}\right \rfloor \geq I$ $\endgroup$ – Jean Marie May 20 at 11:48

$$\left\lfloor\frac{AB}{C}\right\rfloor \ge I$$ $$\implies \frac{AB}{C} \ge I$$

$$\implies A \ge \frac{IC}{B}$$

  • $\begingroup$ This is not an answer : you don't take into account the floor signs. $\endgroup$ – Jean Marie May 20 at 11:46
  • $\begingroup$ Well the floor of a number $\ge$ some integer which implies that the number must be $\ge$ that integer. I skipped the first line. $\endgroup$ – Vizag May 20 at 11:47
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    $\begingroup$ Okay I see your point. But it doesn't seem very likely that you can pin down on a value of $A$ from just knowing the result of an inequality. For ex, and I am assuming I have used the definition of floor correctly, $AB/C \ge I$. What better can be said about $A$ then $A \ge IC/B$? $\endgroup$ – Vizag May 20 at 11:52
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    $\begingroup$ In fact you are right. I beg your pardon for having had a doubt. The minimal value $A=\dfrac{IC}{B}$ when plugged into into relationship $\lfloor{AB/C}\rfloor \geq I$, one gets $\lfloor{I}\rfloor \geq I$ which is truebecause $I$ is an integer... which is the case. * $\endgroup$ – Jean Marie May 20 at 12:05
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    $\begingroup$ You embarrass me by apologizing to me, professor. $\endgroup$ – Vizag May 20 at 14:29

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