Question 572484: three consecutive numbers whose product is 40 times their sum.
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Let a-1, a, a+1 be the numbers (I chose these instead of a, a+1, a+2 for simplicity). Their sum is 3a, so
40(3a) = (a-1)(a)(a+1)
a = 0 works. If a is not zero, then divide through by a:
120 = (a-1)(a+1) = a^2 - 1
Therefore a^2 = 121, a = 11 or -11. The numbers are either {-1,0,1}, {10,11,12} or {-12,-11,-10}.
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