Let $ M=\{1,2,...,2013\} $ and let $ \Gamma $ be a circle. For every nonempty subset $ B $ of the set $ M $, denote by $ S(B) $ sum of elements of the set $ B $, and define $ S(\varnothing)=0 $ ( $ \varnothing $ is the empty set ). Is it possible to join every subset $ B $ of $ M $ with some point $ A $ on the circle $ \Gamma $ so that following conditions are fulfilled: $ 1 $. Different subsets are joined with different points; $ 2 $. All joined points are vertices of a regular polygon; $ 3 $. If $ A_1,A_2,...,A_k $ are some of the joined points, $ k>2 $ , such that $ A_1A_2...A_k $ is a regular $ k-gon $, then $ 2014 $ divides $ S(B_1)+S(B_2)+...+S(B_k) $ ?
Problem
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Tags: combinatorics
Naysh
16.05.2015 22:36
Is there any condition on which points are connected, or how many points are connected (since we could otherwise just not connect any points)?
arberig
16.05.2015 22:48
Naysh wrote: Is there any condition on which points are connected, or how many points are connected (since we could otherwise just not connect any points)? No there isn't any other condition !
lian_the_noob12
04.08.2024 20:09
The answer is $\textbf{YES}$ I don't know how to write up this solution. Here, only $2^{l}-gon$ is possible, So, we need to partition the subsets (or $\textbf{S(B)}$) into $2^{2011}$ groups having sum $=\frac{1}{2^{2011}} (1+2+...2013)\times 2^{2012}=2013\times 2014$ for each so the Condition $3$ satisfies for regular $4-gon$ If it satisfies for $4-gon$ then satisfies for any $2^l, l \ge 3$ gon as we can just join $2,4,8,..$ partitions to make $\frac{1}{2},\frac{1}{4},..$ of the groups which satisfy condition $3$ There exist a partition for $4-gon$ but I don't know how to generalize it for so big number $2^{2013}$