***SPOILER WARNING*** I wrote: << For some integer n > 0 you have all 2^n vectors of dimension n whose entries are +1 or -1. Of course the sum of all 2^n of these vectors is the 0 vector. Then your three-year-old child changes some of the entries of some of these vectors to 0. Show that there is still a nonempty subset of the new set of 2^n vectors that sums to the 0 vector.

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A solution to the puzzle, due to Alon Amit, is described 40 lines below. --Dan ------------------------------------------------------------------------------------------- Here "modified vector" refers to a vector in the set of vectors after it has been somewhat mangled by the three-year-old. (Any of these may of course be unmodified vectors.) We inductively construct a sequence {v(k)} of n-vectors lying in {0,1}^n. Let v(0) be the modified vector that started life as all 1's. Given v(k), let u(k) be the modified vector that started life with -1's precisely in the same places that v(k) has its 1's. Now define: v(k+1) := v(k) + u(k). Then v(k+1) must also lie in {0,1}^n. This gives an infinite sequence of v(k) in {0,1}^n. Since {0,1}^n is finite, we must have v(k) = v(k+r) for some k >=0, r >= 1. Then v(k+r) - v(k) is a sum of modified vectors that sums to the 0 vector. ------------------------------------------------------------------------------------------- ________________________________________________________________________________________ It goes without saying that .