I got the following from a book:
Let . Show that for every set S of integers, there is a nonempty subset of such that divides the sum of elements of .
Let for each integer with . For each integer
for some integer , where .
We consider two cases:
for some integer . Then , that is, divides the sum of the elements .
for all integer .
Hence there exist integers there exist integers and with such that and for integer with .
1. I have tested a set of elements on the computer and found that at least one of the nonempty subsets is divisible by .
2. In regard to case 2 in the proof above, I can't see how
implies the existence of and . It might be true by chance.
Question: Does anyone here have a better proof than this?