How can I solve an equation with multiple floor functions added together?

$$ 18 + \lfloor 2.6 \rfloor + \lfloor x \rfloor + \left\lfloor \frac x4 \right\rfloor + 5 = 1 $$


closed as off-topic by Saad, steven gregory, M. Vinay, Xander Henderson, Alex Provost Mar 14 at 19:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, steven gregory, M. Vinay, Xander Henderson, Alex Provost
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Welcome to Math Stack Exchange. Floor(2.6)=2, right? $\endgroup$ – J. W. Tanner Mar 14 at 12:46

$\lfloor x\rfloor+\lfloor x/4\rfloor=-24$. Use the fact $m-1<\lfloor m\rfloor\le m$ to conclude $5x/4-2<-24\le5x/4$. Solve these simultaneous inequalities to narrow down the search to the interval $[-19.2,-17.6)$. Note that $\lfloor x/4\rfloor=-5$ in the entire interval, so you need the subset where $\lfloor x\rfloor=-19$, i.e.$[-19,-18)$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.