$31 \cdot 26$ satisfies the equation $m^{3}+6 m^{2}+5 m=27 n^{3}+9 n^{2}+9 n+1$ for the number of integer pairs $(m, n)$ is
(A) 0.
(B) 1.
(C) 3.
(D) 9.
(E) infinitely many.
(30th American High School Mathematics Examination, 1979)
[Solution] Since $m^{3}+6 m^{2}+5 m=m(m+1)(m+5)$, for all integers $m$, it can be divisible by 3, which only requires noting that when the remainder of $m$ divided by 3 is 0, 1, or 2, $m$, $m+5$, or $m+1$ can be divisible by 3, respectively.
However, $27 n^{3}+9 n^{2}+9 n+1$ always leaves a remainder of 1 when divided by 3.
Thus, there are no integer pairs $(m, n)$ that satisfy the given equation.
Therefore, the answer is $(A)$.