1951 AHSME Problems/Problem 15

Problem

The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$, is:

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$

Solution 1

Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3. Therefore $6$ must divide the given expression. Plugging in $n=2$ yields $6$. So the largest possibility is $6$.

Clearly the answer is $\boxed{\textbf{(E)} \ 6}$

Solution 2

In general, $r!$ | $n(n+1)(n+2)...(n+r-1)$ were $r$ and $n$ are integers. So here $3!$ | $n^3$ - $n$ always for any integer $n$.Hence,the correct answer is $6$. $\boxed{\textbf{(E)} \ 6}$

~geometry wizard.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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