Question 658206
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No solution.  The sum of an odd number of odd numbers is odd, and 52 is even.


Proof:  *[tex \LARGE 2n\ -\ 1] is the *[tex \LARGE n\text{th}] odd number where *[tex \LARGE n\ \in\ \mathbb{Z}], hence 5 arbitrary odd numbers can be represented as *[tex \LARGE 2n_1\ -\ 1], *[tex \LARGE 2n_2\ -\ 1], *[tex \LARGE 2n_3\ -\ 1], *[tex \LARGE 2n_4\ -\ 1], and *[tex \LARGE 2n_5\ -\ 1].  And their sum would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2n_1\ -\ 1\ +\ 2n_2\ -\ 1\ +\ 2n_3\ -\ 1\ +\ 2n_4\ -\ 1\ +\ 2n_5\ -\ 1]


Which can be rearranged thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(n_1\ +\ n_2\ +\ n_3\ +\ n_4\ +\ n_5\right)\ -\ 5]


Which can clearly be seen is not evenly divisible by 2.  Therefore the sum of any five odd integers is odd.   The proof can easily be generalized to cover any odd number of odd numbers.



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