Question 986585
<font face="Times New Roman" size="+2">


Let *[tex \Large x] represent the smallest of the three consecutive odd integers. Since this odd integer is arbitrary, at least so far, we can say that it is the *[tex \Large n]th odd integer, hence *[tex \Large x\ =\ 2n\ +\ 1].  The next consecutive odd integer must then be the *[tex \Large n\ +\ 1]th odd integer, or *[tex \Large 2(n\ +\ 1)\ +\ 1], which is to say *[tex \Large 2n\ +\ 3].  Since *[tex \Large x\ =\ 2n\ +\ 1], *[tex \Large 2n\ +\ 3] must be *[tex \Large x\ +\ 2].  Similarly, the next consecutive odd integer is *[tex \Large x\ +\ 4]


Since the sum of these three odd integers is 63:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ (x\ +\ 2)\ +\ (x\ +\ 4)\ =\ 63]


Solve for *[tex \Large x] then compute *[tex \Large x\ +\ 2] and *[tex \Large x\ +\ 4]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \