Question 903078
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No.

Let *[tex \Large x_e] represent an arbitrary even integer.  Then *[tex \Large x_e\ +\ 2] is the next consecutive even integer.  *[tex \Large x_e\ +\ 4] is the next after that, and so on.  The sum of five consecutive even integers is then *[tex \Large 5x_e\ +\ 20], and the average is then that sum divided by 5, namely *[tex \Large x_e\ +\ 4].


Similarly, if *[tex \Large x_o] is an arbitrary odd integer, the sum of five consecutive odd integers would be *[tex \Large 5x_o\ +\ 20] and the average would be *[tex \Large x_o\ +\ 4].


In order for there to be a set of five consecutive even integers that have an average equal to the average of a set of five consecutive odd integers, then the statement:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x_e\ +\ 4\ =\ x_o\ +\ 4]


must be true and therefore


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x_e\ =\ x_o]


must be true.


Since there is no even integer that is equal to an odd integer, this relation can never be true.  Hence, a set of 5 consecutive even integers whose average is equal to the average of a set of 5 consecutive odd integers does not exist.
  

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