$33 \cdot 20 n$ is a natural number, when substituting into the algebraic expression $n^{3}-n$ to calculate its value, four students came up with the following four results, the correct result can only be
(A) 388944.
(B) 388945.
(C) 388954.
(D) 388948.
(China Guangzhou, Wuhan, Fuzhou and other five cities junior high school mathematics league, 1991)
[Solution 1] We only need to consider the last digit of the calculation result (the first five digits are the same except for (C)), i.e., consider the influence of the last digit of $n$ on the result:
\begin{tabular}{c|cccccccccc}
\hline Last digit of $n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline Last digit of $n^{3}$ & 0 & 1 & 8 & 7 & 4 & 5 & 6 & 3 & 2 & 9 \\
\hline Last digit of $n^{3}-n$ & 0 & 0 & 6 & 4 & 0 & 0 & 0 & 6 & 4 & 0 \\
\hline
\end{tabular}

From the table above, we can see that the last digit of $n^{3}-n$ can only be $0, 4, 6$.
Clearly, the correct answer can only be (A) or (C).
Furthermore, since $n^{3}-n=(n-1) n(n+1)$, it is divisible by 3. A number divisible by 3 has the sum of its digits as a multiple of 3.
Notice that $3+8+8+9+4+4=36$, while
$$
3+8+8+9+5+4=37.
$$

Therefore, the correct choice is (A).
[Solution 2] Since $n^{3}-n=(n-1) n(n+1)$, then $6 \mid (n^{3}-n)$.
And $6 \mid 388944$,
while $388954=388944+10$, and $388948=388944+4$,
so $6 \nmid 388954, 6 \nmid 388944$.
Also, 388945 is an odd number, so $6 \nmid 388945$.
In summary, the correct result can only be 388944.
Therefore, the correct choice is (A).