$31 \cdot 27$ satisfies $m n \geqslant 0$ and $m^{3}+n^{3}+99 m n=33^{3}$ for the number of integer pairs $(m, n)$ is equal to
(A) 2.
(B) 3.
(C) 33.
(D) 35.
(E) 99.
(50th American High School Mathematics Examination, 1999)
[Solution] From $(m+n)^{3}=m^{3}+3 m n(m+n)+n^{3}$, it is clear that when $m+n=33$, $m^{3}+n^{3}+99 m n=33^{3}$. Given the condition $m n \geqslant 0$, there are 34 pairs of $(m, n)$: $(0,33),(1,32), \cdots,(32,1),(33,0)$. Additionally, when $m=-33, n=-33$, the equation is also satisfied. In total, there are 35 pairs of $(m, n)$ that satisfy the condition. Therefore, the answer is $(D)$.