Question 286002
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Let *[tex \Large x] represent the smallest of the three integers.  Then then next odd integer must be *[tex \Large x\ +\ 2], and the one after that has to be *[tex \Large x\ +\ 4]


The sum of the three is


*[tex \LARGE\ \ \ \ \ \ \ \ \ \  x\ +\ x\ +\ 2\ +\ x\ +\ 4\ =\ 3x\ +\ 6]


Two times the sum is then:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ 6x\ +\ 12]


The product of the first and second is:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x]


So, putting it all together:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \  6x\ +\ 12\ =\ x^2\ +\ 2x\ -\ 153]


Standard form:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x^2\ -\ 4x\ - \ 165\ =\ 0]


Solve the factorable quadratic.  Exclude the negative root because the problem asks for positive integers.


John
*[tex \Large e^{i\pi}\ +\ 1\ =\ 0]
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