1956 AHSME Problems/Problem 34

Problem 34

If $n$ is any whole number, $n^2(n^2 - 1)$ is always divisible by

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ \text{any multiple of }12\qquad \textbf{(D)}\ 12-n\qquad \textbf{(E)}\ 12\text{ and }24$

Solution

Suppose $n$ is even. So, we have $n^2(n+1)(n-1).$ Out of these three numbers, at least one of them is going to be a multiple of 3. $n^2$ is also a multiple of 4. Therefore, this expression is always divisible by $\boxed{\textbf{(A)} \quad 12}.$

-coolmath34

1956 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 33	Followed by Problem 35
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1956 AHSME Problems/Problem 34

Problem 34

Solution

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