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 5 of 6: Count both sides of the injection
mm≤(m(m−1)+1)∣A∣m^m\le(m(m-1)+1)^{|A|}
Detailed analysis

There are exactly mmm^m multiples of mm in [0,mm+1)[0,m^{m+1}), namely X=0,m,2m,…,(mm−1)mX=0,m,2m,\ldots,(m^m-1)m, and each maps injectively into the set of ∣A∣|A|-tuples with entries in {0,1,…,m(m−1)}\{0,1,\ldots,m(m-1)\}, which has (m(m−1)+1)∣A∣(m(m-1)+1)^{|A|} elements. Hence mm≤(m(m−1)+1)∣A∣m^m\le(m(m-1)+1)^{|A|}.