2010/11 AHSMC Part 1 Problem 16

When $n^3 - 3n + 2$ is divided by $2n + 1$, the quotient is $\frac{n^2}{2} - \frac{n}{4} - \frac{11}{8}$ and the remainder is $\frac{27}{8}$. Hence $8(n^3 - 3n + 2) = (2n+1)(4n^2 - 2n - 11) + 27$, so that $2n+1$ divides $n^3 - 3n + 2$ if and only if it divides $27$. The set of all values of $n$ for which $2n+1$ divides $27$ is $\{-14, -5, -2, -1, 0, 1, 4, 13\}$, and there are $8$ such values. The answer is (E).