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.

  • $\begingroup$ 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. $\endgroup$ – Stan Tendijck May 16 at 14:12
  • $\begingroup$ @StanTendijck hmmm... I'll try. $\endgroup$ – 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!

  • $\begingroup$ Thanks! That solved it! $\endgroup$ – PranavGupta53535 May 16 at 14:32
  • $\begingroup$ Upvote the question, if u will. $\endgroup$ – PranavGupta53535 May 16 at 14:32

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