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.  
<pre>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
{{{5(3S + 6) = S^2 + 2S + 66}}}
{{{15S + 30 = S^2 + 2S + 66}}}
{{{S^2 + 2S - 15S + 66 - 30 = 0}}}
{{{S^2 - 13S + 36 = 0}}}
(S - 9)(S - 4) = 0
S, or smallest integer = {{{highlight_green(9)}}}            OR          S = 4 (ignore)
Middle integer: 9 + 2, or {{{highlight_green(11)}}}
Largest integer: 9 + 4, or {{{highlight_green(13)}}}