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 Choosing the right number (Posted on 2018-02-13)
From a set of 2n+1 consecutive integers a certain number, say x, is selected in a way that the remaining n pairs can be arranged as couples having the same sum.

What are all the possible values of x making such task doable?

 See The Solution Submitted by Ady TZIDON No Rating

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 Not feeling literal | Comment 2 of 6 |
Literally, the only way to have n pairs is to remove 1 from 2n+1.

However, I suspect that Ady really wants us to have (2n + 1 - x)/2 pairs.

In this case, x can be any odd number from 1 to 2n-3.  The x numbers can be removed from the low end of the sequence, or from the high end, or from the exact middle.

If x = 2n-1, then there is only one remaining pair,  and arguably the puzzle condition is not met.

If x = 2n+1, then there are no remaining pairs,  and arguably the puzzle condition is not met.

An earlier solution suggested n + 1, but this fails if n is odd.

 Posted by Steve Herman on 2018-02-13 20:24:52

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