Question 321687
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Let *[tex \Large x\ +\ i] represent the 7 integers where *[tex \Large 0\ \leq\ i\ \leq\ 6].


The sum of the 7 integers is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,0}^6\ x\ +\ i]


where *[tex \LARGE x] is an arbitrary positive integer.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,0}^6\ x\ +\ i\ =\ 7x\ +\ 21]


If *[tex \LARGE x] is the *[tex \LARGE n]th even integer, then *[tex \LARGE x\ =\ 2n], therefore *[tex \LARGE 7x\ =\ 14n] and is even, hence *[tex \LARGE 7x\ +\ 21] is odd.


If *[tex \LARGE x] is the *[tex \LARGE n]th odd integer, then *[tex \LARGE x\ =\ 2n\ -\ 1], therefore *[tex \LARGE 7x\ =\ 13n] and is odd, hence *[tex \LARGE 7x\ +\ 21] is even.


Which is to say that answers A, C, and D can be excluded.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7x\ +\ 21]


is only divisible by 3 if *[tex \Large x] is divisible by 3.  Exclude answer E.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{7x\ +\ 21}{7}\ =\ x\ +\ 3]


Evenly divisible by 7 regardless of the value of x.  Answer B.





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