33.24 If $[a]$ represents the integer part of the real number $a$, then $\left[\frac{1}{\sqrt{16-6 \sqrt{7}}}\right]$ equals
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(Shaanxi Province, China Junior High School Mathematics Competition, 1997)
[Solution] $\because 16-6 \sqrt{7}=9+7-2 \sqrt{63}=(\sqrt{9}-\sqrt{7})^{2}$, $\therefore \quad \frac{1}{\sqrt{16-6 \sqrt{7}}}=\frac{1}{\sqrt{9}-\sqrt{7}}=\frac{3+\sqrt{7}}{2}$.
And $2<\sqrt{7}<3$,
$$
\therefore \quad \frac{5}{2}<\frac{3+\sqrt{7}}{2}<3 \text {. }
$$

It is clear that $\left[\frac{1}{\sqrt{16-6 \sqrt{7}}}\right]=2$.
Therefore, the answer is $(B)$.