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.
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!