SOLUTION: Find three consecutive odd positive intergers such that 5 times the sum of all three is 66 more then the product of the first and second intergers.

Question 380930: Find three consecutive odd positive intergers such that 5 times the sum of all three is 66 more then the product of the first and second intergers.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let the integers be x,x+2,x+4
...
5(x+x+2+x+4)=x(x+2)+66
5(3x+6)=x^2+2x+66
15x+30=x^2+2x+66
-15x
30=x^2+2x-15x+66
30=x^2-13x+66
-30
0=x^2-13x+36
x^2-9x-4x+36=0
x(x-9)-4(x-9)=0
(x-9)(x-4)=0
x= 9 OR 4.
reject 4
x= 9
the numbers are 9,11,13.
...
m.ananth@hotmail.ca

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