$35 \cdot 8$ A wooden cube has a side length of $n$ units (where $n>2$ is an integer). The entire surface of the cube is painted black, and then it is cut into $n^{3}$ smaller cubes, each with a side length of 1 unit, using planes parallel to the faces of the original cube. If the number of smaller cubes with exactly one face painted black is equal to the number of smaller cubes with no faces painted, $n$ is
(A) 5 .
(B) 6 .
(C) 7
(D) 8
(E) None of the above
(36th American High School Mathematics Examination, 1985)
[Solution] According to the problem, the number of small cubes with exactly one face painted is $6 \cdot(n-2)^{2}$, and the number of small cubes with no faces painted is $(n-2)^{3}$. According to the problem, we have

that is $\square$
$$
\begin{array}{l}
(n-2)^{3}=6(n-2)^{2}, \\
n-2=6 \quad(\because n>2) .
\end{array}
$$

Therefore, $n=8$.
So the answer is $(D)$.