Question 335536
<font face="Garamond" size="+2">


Let *[tex \Large x] represent the least of the three consecutive integers.  The second one is then *[tex \Large x\ +\ 1] and the next one is *[tex \Large x\ +\ 2]


Twice the third:  *[tex \Large 2(x\ +\ 2)\ =\ 2x\ +\ 4]


Three times the sum of the first and third plus 13:  *[tex \Large 3(x\ +\ x\ +\ 1)\ +\ 13\ =\ 6x\ +\ 16]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x\ +\ 16\ =\ 2x\ +\ 4]


Solve for *[tex \Large x], calculate the other two integers by adding 1 and then adding 1 again.  Multiply the three together, then divide by -2.  Let me know how you did.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
</font>