Competition problem — Number theory

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Number theory Difficulty 2.4 Junior Find the answer

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

Pick one

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

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.