10. Let $a>1$ be a positive real number, and $n \geqslant 2$ be a natural number, and the equation $[a x]=x$ has exactly $n$ distinct solutions, then the range of values for $a$ is . $\qquad$
10. $\left[1+\frac{1}{n}, 1+\frac{1}{n-1}\right)$

By the problem statement, $x$ must be an integer. When $x<0$, $[a x]<a x<x$, so $x \geqslant 0$. Let $\{a\}=a-[a]$, then the original equation becomes $x=[a x]=[a] x+[\{a\} x]$. Since $[a] \geqslant 1$, the above equation holds $\Leftrightarrow[a]=1$, and $\{a\} x<1$. Also, $x=0$ clearly satisfies the condition.

Therefore, $x<\frac{1}{\{a\}}$ holds exactly for $n-1$ positive integers $1,2, \cdots, n-1$, thus $\frac{1}{n} \leqslant\{a\}<\frac{1}{n-1}$. That is, $1+\frac{1}{n} \leqslant a<1+\frac{1}{n-1}$.