$30 \cdot 43$ A two-digit number $N$ minus the number with its digits reversed, the result is a perfect cube, then
(A) $N$ cannot end in 5.
(B) Except for 5, $N$ can end in any digit.
(C) $N$ does not exist.
(D) $N$ has exactly 7 values.
(E) $N$ has exactly 10 values.
(8th American High School Mathematics Examination, 1957)
［Solution］Let the two-digit number be $10 m+n$, then
$$
10 m+n-(10 n+m)=9(m-n) .
$$

Since $m-n<9$, it follows that $9(m-n)<81$.
Therefore, these perfect cubes can only be $1,8,27,64$, but among them, only 27 is a multiple of 9.
Thus, $9(m-n)=27$. That is, $m=3+n$.
Also, $0<m \leqslant 9$, so $0 \leqslant n \leqslant 6$.
When $n=0,1,2,3,4,5,6$, the two-digit numbers are
$$
30,41,52,63,74,85,96 .
$$

Therefore, the answer is $(D)$.