$37 \cdot 15$  Two spheres, one with its center at $\left(0,0, \frac{21}{2}\right)$ and radius 6; the other with its center at $(0,0,1)$ and radius $\frac{9}{2}$. Let the coordinates of the points of intersection of the two spheres be $(x, y, z)$, the number of lattice points (coordinates all integers) is 
(A) 7.
(B) 9.
(C) 11.
(D) 1.3.
(E) 15.
(47th American High School Mathematics Examination, 1996)
［Solution］From the problem, we know the equations of the two spheres are
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+\left(z-\frac{21}{2}\right)^{2} \leqslant 6^{2}, \\
x^{2}+y^{2}+(z-1)^{2} \leqslant\left(\frac{9}{2}\right)^{2} .
\end{array}\right.
$$

From the above, we have $\frac{9}{2} \leqslant z \leqslant \frac{11}{2}$. Considering the lattice points within this range, then $z=5$.
At this point, we have $x^{2}+y^{2} \leqslant 4-\frac{1}{4}$. If $x, y$ are both integers, then $x^{2}+y^{2} \leqslant 4$. The integer pairs $(x, y)$ that satisfy the above inequality are
$$
\begin{array}{l}
(0,0),(1,1),(-1,1),(1,-1),(-1,-1), \\
(1,0),(0,1),(-1,0),(0,-1) .
\end{array}
$$

There are 9 pairs in total, so the number of lattice points $(x, y, z)$ is 9.
Therefore, the answer is $(B)$.