The sum of three consecutive integers is equal to their product. How many such possibilities are there?

Correct Answer: (C) Only three

Let the three consecutive integers be $n - 1, n, n + 1$. Their sum is $3n$ and their product is $(n - 1)n(n + 1) = n(n^2 - 1)$. Setting them equal gives $3n = n(n^2 - 1)$$\implies$ n(n^2 - 4) = 0, yielding solutions $n = 0$, $n = 2$, and $n = -2$. The corresponding sets of integers are: (1) $\{-1, 0, 1\}$ with sum 0 and product 0; (2) $\{1, 2, 3\}$ with sum 6 and product 6; and (3) $\{-3, -2, -1\}$ with sum -6 and product -6. Thus, there are exactly three such possibilities.