# Can anyone help me for this first degree floor function equation?

Find $$y$$ such that

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

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

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

and

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

But I’m stuck, could anyone help me?

• 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). Nov 14, 2018 at 15:00
• @HavanaTime Please do not vandalize the post after you have received an answer. Doing so can get you into trouble on this site.
– user279515
Dec 15, 2018 at 4:40

Hint: write y as $$n+d$$ .Now the first term turns out to be n and for second term take cases for $$0 ,$$1/3<=d<2/3$$ and $$2/3<=d<1$$. Can you do after that ?
• answer should be $(4/3,5/3)$ Nov 14, 2018 at 15:38
Hint: $$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.