Question 203845
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The total number of candies can be any even number greater than 2.


Let *[tex \Large x] represent the number of candies below the red one.  Then *[tex \Large x - 1] must represent the number of candies above the red one, and *[tex \Large 2x] must be the total number of candies.


But the total number of candies is also given by adding the number below the red one, the red one, and the number above the red one, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x + 1 + (x - 1) = 2x]


Which, as should be obvious to the most casual observer, is true for all real *[tex \Large x].


Since it is reasonable to presume that we are counting whole candies, we can restrict our investigation to the positive integers.  Since we cannot have a negative number of candies above the red one, the smallest value *[tex \Large x] can assume is 1 (making the number of candies above the red one be 0 which is 1 less than 1 and still fits the given conditions), hence the smallest total we can have is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (1) + 1 + ((1) - 1) = 2(1) = 2]


There is no upper bound on *[tex \Large x], hence the total can be as large as we like by taking a sufficiently large *[tex \Large x].  But we know that *[tex \Large 2x] is divisible by 2, therefore the total must be an even integer.


That's the best I can do unless you left something out of the problem statement.


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