Each of ten people around a circle chooses a number and tells it to the neighbor on each side. Thus each person gives out one
number and receives two numbers. The players then announce the average of the two numbers they received. The announced
numbers, in order around the circle were $$1,2,3,4,5,6,7,8,9,10$$. The number chosen by the person who announced the number $$6$$ is
Solution
Let the chosen numbers be $$x_1,x_2,\ldots,x_{10}$$ cyclically. The announcements give $$x_{i-1}+x_{i+1}=2i$$ for every $$i$$. Solving these cyclic equations gives $$(x_1,\ldots,x_{10})=(6,-3,-2,9,10,1,2,13,14,5)$$, so the person announcing $$6$$ chose $$1$$.
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