Question

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

• A. 12
• B. 24
• C. any multiple of 12
• D. 12-n
• E. 12 and 24

Answer 1

The correct option is A 12

By considering the special case n = 2 we immediately rule out choices (b), (c), (d), and (e). We now show that (a) holds.
$n^2(n^2-1)=n(n-1)n(n+1)$= nk,where k is a product of three consecutive integers, hence always divisible by 3. If n is even, k is divisible also by 2 (since n is a factor of k), hence by 6,and nk by 12; if n is odd, k is divisible also by 4 (since the even numbers n- 1 and n + 1 are factors of k), hence by 12.