Problem:
Given a sequence of real numbers, a move consists of choosing two terms and replacing each by their arithmetic mean. Show that there exists a sequence of 2015 distinct real numbers such that after one initial move is applied to the sequence - no matter what move - there is always a way to continue with a finite sequence of moves so as to obtain in the end a constant sequence.
Solution:
The sequence (x1​,x2​,…,x2015​)=(1,2,…,2015) satisfies the required property (as does any arithmetic sequence).
Assume that (xm​,xn​)=(m,n) is replaced by (2m+n​,2m+n​) in the first move. We consider two cases.
In the first case, we assume that none of m and n is equal to 1008 . In the second move, we replace (x2016−m​,x2016−n​)=(2016−m,2016−n) by (2016−2m+n​,2016−2m+n​). Let all the subsequent moves be applied to the pairs (xj​,x2016−j​),j=1,2,…,1008. This yields the constant sequence (1008,1008,…,1008).
In the second case, we assume that one of m and n, say, n is equal to 1008 . After the first move we have xm​=x1008​=21008+m​. Choose k different from 1008,m, and 2016−m. We illustrate our next four moves in the following table. (In each move, we operate on the the numbers in bold.)
(xk​,xm​,x1008​,x2016−m​,x2016−k​)=(k,21008+m​,21008+m​,2016−m,2016−k)→(1008,21008+m​,21008+m​,2016−m,1008)→(23024−m​,21008+m​,21008+m​,23024−m​,1008)→(1008,1008,21008+m​,23024−m​,1008)→(1008,1008,1008,1008,1008)​
Finally apply the move to all the pairs (xj​,x2016−j​) (with jî€ =m,k,2016−m,2016−k ) to obtain the constant sequence (1008,1008,…,1008).
Query: If the initial sequence is (1,2,3,…,2013,2014,2016), where " 2015 " is replaced by "2016", is it possible to obtain a constant sequence after a finite sequence of moves?
The problems on this page are the property of the MAA's American Mathematics Competitions