Discussion
If n is a whole number greater than 1, then \(n^2(n^2-1)\) is always divisible by:
Answer: A
The expression can be factored as \(n^2(n-1)(n+1)\), or \(n \times n \times (n-1) \times (n+1)\). This can be rearranged as \((n-1)n(n+1)n\). The term \((n-1)n(n+1)\) is the product of three consecutive integers, which is always divisible by \(3! = 6\). So the expression is divisible by 6. We also need to check for divisibility by other factors. The expression is divisible by 3 and 2. We need to check for 4. If n is even, let n=2k. Then \(n^2=4k^2\), so the expression is divisible by 4. If n is odd, let n=2k+1. Then n-1 and n+1 are consecutive even numbers, \((2k)(2k+2) = 4k(k+1)\). This product is divisible by 4. So the expression is always divisible by 3 and 4, which are co-prime. Therefore, it is always divisible by \(3 \times 4 = 12\).
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