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 2 of 6: Reindex the sum over the elements of AA
X=∑a∈Afa(X) a,fa(X)=∑i: a∈BiciX=\sum_{a\in A}f_a(X)\,a,\quad f_a(X)=\sum_{i:\,a\in B_i}c_i
Detailed analysis

Substituting mi=∑b∈Bibm^i=\sum_{b\in B_i}b and swapping the order of summation, X=∑i=1mci∑b∈Bib=∑a∈A(∑i: a∈Bici)a=∑a∈Afa(X) aX=\sum_{i=1}^m c_i\sum_{b\in B_i}b=\sum_{a\in A}\Big(\sum_{i:\,a\in B_i}c_i\Big)a=\sum_{a\in A}f_a(X)\,a, where fa(X):=∑i: a∈Bicif_a(X):=\sum_{i:\,a\in B_i}c_i.