olymmath

RU-ALLRUSSIAN 1976 Problem 14

    You are given a regular n- gon. Each vertex is marked +1+1 or - 1. A move consists of changing the sign of all the vertices which form a regular k- gon for some 1<k≪=n1 < k \ll = n . [A regular 2- gon means two vertices which have the center of the n- gon as their midpoint.]. For example, if we label the vertices of a regular 6- gon 1, 2, 3, 4, 5, 6, then you can change the sign of {1, 4}, {2, 5}, {3, 6}, {1, 3, 5}, {2, 4, 6} or {1, 2, 3, 4, 5, 6}. Show that for (1) n=15n = 15 , (2) n=30n = 30 , (3) any n>2n > 2 , we can find some initial marking which cannot be changed to all +1+1 by any series of

    moves. Let f(n) be the largest number of markings, so that no one can be obtained from any other by any series of moves. Show that f(200)=280\mathrm{f}(200) = 2^{80} .

    Dataset text NuminaMath olympiads. Quoted with source.

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