$29 \cdot 21$ If $n$ is any integer, $n^{2}\left(n^{2}-1\right)$ is divisible by $x$, then $x$ equals
(A) 12.
(B) 24.
(C) Any multiple of 12.
(D) $12-n$.
(E) 12 and 24.
(7th American High School Mathematics Examination, 1956)
[Solution] When $n=2$, $n^{2}\left(n^{2}-1\right)=12$, so $(B),(C),(D),(E)$ are not true. Therefore, the answer is $(A)$.