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Essentially it is to be proven that
3^2016 + 4^2017 > (2018^183)^2
3^2016 + 4^2017 > 2018^366
4^2017 = 2^4034
2048 = 2^11
2048^366 = 2^(11*366) = 2^4026
(4^2017 = 2^4034) > (2048^366 = 2^4026) > 2018^366
4^2017 > 2018^366
So its only logical that
4^2017 + 3^2016 > 2018^366.
Is it possible to write down 1,2,3..100 in some order (one after an other), such that the sum of any two adjacent numbers is a prime number?
In a sequence of eleven real numbers, only two are shown. The product of every 3 successive boxes is 120. Find the sum of all the numbers in the boxes (including the two already open).
_ _ (6) _ _ _ _ _ _ (-4) _
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