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 3 of 6: Bound fa(X)f_a(X) for every aa and XX
0≤fa(X)≤m(m−1)0\le f_a(X)\le m(m-1)
Detailed analysis

Since fa(X)f_a(X) is a sum of at most mm digits, each in {0,1,…,m−1}\{0,1,\ldots,m-1\}, we get 0≤fa(X)≤m(m−1)0\le f_a(X)\le m(m-1) for every a∈Aa\in A and every multiple XX of mm below mm+1m^{m+1}.