$33 \cdot 23$ Let $[x]$ denote the greatest integer not greater than $x$. Given $\left[\frac{\left[\frac{91}{n}\right]}{n}\right]=1$, where $n$ is a positive integer. Then $n$ is
(A) $7,8,9$.
(B) $7,8,9,10$.
(C) $6,7,8,9$.
(D) 7,8 .
(3rd "Five Sheep Cup" Junior High School Mathematics Competition, 1991)
[Solution] When $n$ is a positive integer, $x=\left[\frac{\left[\frac{91}{n}\right]}{n}\right]$ does not increase (decreases or remains unchanged) as $n$ increases.
Also, $\left[\frac{\frac{91}{n}}{n}\right] \leqslant\left[\frac{\frac{91}{n}}{n}\right]=\left[\frac{91}{n^{2}}\right]$.
If $\left[\frac{91}{n^{2}}\right] \geqslant 1$, then the maximum value of $n$ is 9.
It is easy to calculate that when $n=9,8,7,6$, $\quad x=1,1,1,2$.
Therefore, when $n=7,8,9$, these are the solutions.
Hence, the answer is $(A)$.