The sum of three consecutive integers is equal to their product. How many such possibilities are there?
The sum of three consecutive integers is equal to their product. How many such possibilities are there?
Only one
Only two
Only three
No such possibility is there
Solution
Let the three consecutive integers be $x - 1$, $x$, $x + 1$. Then $(x-1) + x + (x+1) = (x-1) \cdot x \cdot (x+1)$, i.e. $3x = x(x^2 - 1)$, so $x^3 - 4x = 0$ and $x(x^2 - 4) = 0$, giving $x = 0, 2, -2$. Hence there are three possibilities: (-1,0,1), (1,2,3) and (-3,-2,-1).