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f(A,B) - Iran NMO 1998 (Second Round) Problem6

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October 4, 2010
combinatorics proposedcombinatorics

Problem Statement

If A=(a1,⋯ ,an)A=(a_1,\cdots,a_n) , B=(b1,⋯ ,bn)B=(b_1,\cdots,b_n) be 22 n−n-tuple that ai,bi=0 or 1a_i,b_i=0 \ or \ 1 for i=1,2,⋯ ,ni=1,2,\cdots,n, we define f(A,B)f(A,B) the number of 1≤i≤n1\leq i\leq n that ai≠bia_i\ne b_i. For instance, if A=(0,1,1)A=(0,1,1) , B=(1,1,0)B=(1,1,0), then f(A,B)=2f(A,B)=2. Now, let A=(a1,⋯ ,an)A=(a_1,\cdots,a_n) , B=(b1,⋯ ,bn)B=(b_1,\cdots,b_n) , C=(c1,⋯ ,cn)C=(c_1,\cdots,c_n) be 3 n−n-tuple, such that for i=1,2,⋯ ,ni=1,2,\cdots,n, ai,bi,ci=0 or 1a_i,b_i,c_i=0 \ or \ 1 and f(A,B)=f(A,C)=f(B,C)=df(A,B)=f(A,C)=f(B,C)=d. a)a) Prove that dd is even. b)b) Prove that there exists a n−n-tuple D=(d1,⋯ ,dn)D=(d_1,\cdots,d_n) that di=0 or 1d_i=0 \ or \ 1 for i=1,2,⋯ ,ni=1,2,\cdots,n, such that f(A,D)=f(B,D)=f(C,D)=d2f(A,D)=f(B,D)=f(C,D)=\frac{d}{2}.
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