$29 \cdot 20$ Regardless of the integer value of $n$, the largest number that can divide the expression $n^{3}-n$ is
(A) 2 .
(B) 3 .
(C) 4 .
(D) 5 .
(E) 6 .
(2nd American High School Mathematics Examination, 1951)
[Solution] Since $n^{3}-n=(n-1) n(n+1)$, this is the product of three consecutive integers, which must include one even number and one multiple of 3, so it can be divisible by 6.
$$
\text { Also, when } n=2 \text {, } n^{3}-n=6 \text {. }
$$

Therefore, for any integer $n$, the greatest divisor of $n^{3}-n$ is 6. Hence, the answer is $(E)$.