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

Let the three consecutive integers be n-1, n, and n+1.

1. Calculate their sum: Sum = (n-1) + n + (n+1) = 3n

2. Calculate their product: Product = (n-1) * n * (n+1) = n(n^2 - 1)

3. Set the sum equal to the product: 3n = n(n^2 - 1)

4. Solve the equation for n: We can rearrange the equation: n(n^2 - 1) - 3n = 0 n( (n^2 - 1) - 3 ) = 0 n(n^2 - 4) = 0

   This equation gives us three possible values for n: a) n = 0 b) n^2 - 4 = 0 => n^2 = 4 => n = +2 or n = -2

5. Determine the sets of integers for each value of n: a) If n = 0, the integers are (0-1), 0, (0+1) = -1, 0, 1. Sum = -1 + 0 + 1 = 0 Product = (-1) * 0 * 1 = 0 This is a valid possibility.

   b) If n = 2, the integers are (2-1), 2, (2+1) = 1, 2, 3. Sum = 1 + 2 + 3 = 6 Product = 1 * 2 * 3 = 6 This is a valid possibility.

   c) If n = -2, the integers are (-2-1), -2, (-2+1) = -3, -2, -1. Sum = -3 + (-2) + (-1) = -6 Product = (-3) * (-2) * (-1) = -6 This is a valid possibility.

We have found three distinct sets of consecutive integers that satisfy the condition: {-1, 0, 1}, {1, 2, 3}, and {-3, -2, -1}.

Therefore, there are three such possibilities.

The final answer is C) Only three.