Question 782764
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Let *[tex \LARGE x] represent an even number.  Then *[tex \LARGE x\ +\ 2n] where *[tex \LARGE n\ \in\ \mathbb{Z}] is another, possibly different, even number.  Likewise *[tex \LARGE x\ +\ 2m] and *[tex \LARGE x\ +\ 2p] are also even numbers where *[tex \LARGE m, p\ \in\ \mathbb{Z}].  The sum of *[tex \LARGE x\ +\ x\ +\ 2n\ +\ x\ +\ 2m\ +\ x\ +\ 2p\ =\ 4x\ +\ 2(n\ +\ m\ +\ p)] is divisible by 4 if and only if *[tex \LARGE n\ +\ m\ +\ p] is even.


Ergo, Sometimes True.


Examples:  2 + 4 + 6 + 8 = 16, and 16 is divisible by 4.


2 + 4 + 6 + 10 = 18, and 18 is not divisible by 4.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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