$29 \cdot 23$ sequence $1^{5}-1,2^{5}-2,3^{5}-3, \cdots, n^{5}-n, \cdots$. The largest integer that can divide every term in this sequence is
(A) 1 .
(B) 60 .
(C) 15 .
(D) 120 .
(E) 30 .
(8th American High School Mathematics Examination, 1957)
[Solution] From $n^{5}-n=n\left(n^{4}-1\right)=n(n-1)(n+1)\left(n^{2}+1\right)$, note that among $n-1, n, n+1$, there must be one even number and one multiple of 3, so their product is divisible by 6.
Next, we prove that $n^{5}-n$ contains a factor of 5.

If $n$ leaves a remainder of $0, 1, 4$ when divided by 5; then $n-1, n, n+1$ must include a multiple of 5; if $n$ leaves a remainder of 2, 3 when divided by 5, i.e., $n=5m \pm 2$ ($m$ is an integer), then by
$$
n^{2}+1=(5m \pm 2)^{2}+1=25m^{2} \pm 20m+5
$$

we know that $n^{2}+1$ contains a factor of 5.
In summary, $n^{5}-n$ is divisible by 30. Also, $2^{5}-2=30$.
Therefore, the answer is $(E)$.