# Floor function equation with $n$ solutions

The question is —

The equation $$\lfloor ax \rfloor = x$$ has exactly $$n$$ distinct solutions, given that $$n \in \mathbb{N}, n \geqslant 2$$ and $$a \in \mathbb{R}, a > 1$$. Find the range of $$a$$.

My work —

$$\lfloor ax \rfloor = ax - ax + x$$

$$\Longrightarrow (a-1)x = \{ax\}$$

$$\Longrightarrow 0 \leqslant (a-1)x < 1$$

I don't know how to proceed any further to arrive at the range of $$a$$ or how to involve $$n$$ in the solution. Can someone please help me out? Even a subtle hint is highly appreciated.

• I would approach this as follows: try a range of values of $a$ and plot the graphs of $\lfloor ax\rfloor -x$. Then, I don't think it is that difficult to get a feeling of what is happening. Once you got the feeling, a proof is probably easily found. – Stan Tendijck May 16 at 14:12
• @StanTendijck hmmm... I'll try. – PranavGupta53535 May 16 at 14:20

As you have rightly started, we have $$(a-1)x\leq 1$$ and $$x\geq 0$$. Clearly, the solution takes the form $$x=0,1,2,n-1$$. Hence, we get $$(a-1)(n-1) \leq 1$$ and $$(a-1)n > 1$$. Now you can compute the desired range!