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:

Case 1:

for some integer . Then , that is, divides the sum of the elements .

Case 2:

for all integer .

Hence there exist integers there exist integers and with such that and for integer with .

Therefore,

and so

Hence

Remark:

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?