The number of elements in the set $$\{(a, b, c) \mid a, b, c \text{ are three consecutive integers in some order}, a + b + c = abc\}$$ is __________

Correct Answer: 18

Solution

For $$n - 1$$, $$n$$, $$n + 1$$, the condition is $$3n = n(n^2 - 1)$$, i.e. $$n(n^2 - 4) = 0$$, so $$n = 0$$, 2 or $$-2$$. This gives the sets $$\{-1, 0, 1\}$$, $$\{1, 2, 3\}$$ and $$\{-3, -2, -1\}$$. Each can be arranged in $$3! = 6$$ orders, so the set has $$3 \times 6 = 18$$ elements.

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