MathLabs

Problem 6

Let m≥2m\ge2 be an integer, AA a finite set of (not necessarily positive) integers, and B1,B2,B3,…,BmB_1,B_2,B_3,\ldots,B_m subsets of AA. Suppose that for every k=1,2,…,mk=1,2,\ldots,m the sum of the elements of BkB_k is mkm^k. Prove that AA contains at least m/2m/2 elements.
Step 4 of 6: The tuple (fa(X))a∈A(f_a(X))_{a\in A} determines XX
X is determined by (fa(X))a∈AX\ \text{is determined by}\ (f_a(X))_{a\in A}
Detailed analysis

By the second step, X=∑a∈Afa(X) aX=\sum_{a\in A}f_a(X)\,a is recovered exactly from the tuple of values (fa(X))a∈A(f_a(X))_{a\in A}, so distinct multiples XX of mm below mm+1m^{m+1} must give distinct tuples: the map X↦(fa(X))a∈AX\mapsto(f_a(X))_{a\in A} is injective.