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

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Question 969388: Find three consecutive odd positive integers such that 5 times the sum of all three is 66 more than the product of the first and second integers.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39616) (Show Source):
You can put this solution on YOUR website!
n is an integer.
First, 2n+1
Second, 2n+3
Third, 2n+5

That is the description transcribed into an equation.

Simplify and solve for n, and then evaluate the three positive integers.

Answer by MathTherapy(10551) (Show Source):
You can put this solution on YOUR website!
Find three consecutive odd positive integers such that 5 times the sum of all three is 66 more than the product of the first and second integers.

Let the smallest integer be S
Then other 2 are: S + 2, and S + 4
We then get: 5(S + S + 2 + S + 4) = S(S + 2) + 66

(S - 9)(S - 4) = 0
S, or smallest integer = OR S = 4 (ignore)
Middle integer: 9 + 2, or
Largest integer: 9 + 4, or