Find $y$ such that

$$\lfloor y \rfloor + \lfloor 3y \rfloor = 5$$

First, I use the properties that $$n \leq y <n+1$$ And suppose

$$\lfloor y\rfloor = 5-n$$


$$\lfloor 3y\rfloor = n$$

But I’m stuck, could anyone help me?

  • $\begingroup$ Do you just need one example of such a $y$ (in which case you could just try some reasonable guesses until you find one), or do you want a description of all the solutions $y$ (which is also easy, but better done by thinking than by trial and error). $\endgroup$ Nov 14, 2018 at 15:00
  • 5
    $\begingroup$ @HavanaTime Please do not vandalize the post after you have received an answer. Doing so can get you into trouble on this site. $\endgroup$
    – user279515
    Dec 15, 2018 at 4:40

2 Answers 2


Hint: write y as $n+d$ .Now the first term turns out to be n and for second term take cases for $0<d<1/3$ ,$1/3<=d<2/3$ and $2/3<=d<1$. Can you do after that ?

  • $\begingroup$ answer should be $(4/3,5/3)$ $\endgroup$ Nov 14, 2018 at 15:38

Hint: $1$ is too small and $2$ is too big. So any values of $y$ will have to be between $1$ and $2$.

For $y$ between $1$ and $2$, $\lfloor y \rfloor = 1$.

So you just have to pick values that get the $\lfloor 3y \rfloor$ term to come out right.


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