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 1 of 6: Write multiples of mm in base mm
In plain words

Any multiple of mm below mm+1m^{m+1} has a quotient by mm that fits in mm base-mm digits.

X=∑i=1mcimi,ci∈{0,1,…,m−1}X=\sum_{i=1}^m c_i m^i,\quad c_i\in\{0,1,\ldots,m-1\}
Detailed analysis

If 0≤X<mm+10\le X<m^{m+1} is a multiple of mm, then X/m<mmX/m<m^m, so X/mX/m has a base-mm expansion with at most mm digits, giving X=∑i=1mcimiX=\sum_{i=1}^m c_im^i with each digit ci∈{0,1,…,m−1}c_i\in\{0,1,\ldots,m-1\}.