SOLUTION: When I divided the product of 3 consecutive positive integers by their mean, I got 99. What is the smallest of the three numbers?

Question 76831: When I divided the product of 3 consecutive positive integers by their mean, I got 99. What is the smallest of the three numbers?
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
WHEN YOU HAVE THREE CONSECUTIVE NUMBERS SAY 5,6 & 7 THE MEAN OR AVERAGE IS ALWAYS THE MIDDLE NUMBER (5*6*7)/6=210/6=35
BECAUSE WE'VE ELIMINATED THE MIDDLE NUMBER WE HAVE LEFT
X(X+2)=35
X^2+2X-35=0
(X-5)(X+7)=0
X-5=0
X=5 ANSWER
X+2=7 ANSWER
--------------------------------------------
[X*(X+1)*(X+2)]/(X+1)=99
X(X+2)=99
X^2+2X-99=0
(X-9)(X+11)=0
X-9=0
X=9 ANSWER. THE OTHER TWO NUMBERS ARE 9+1=10 & 9+2=11
PROOF
(9*10*11)/10=99
(9*11)=99
99=99
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