This is a question I have been trying to solve for days. I feel as though there MUST be some application of the pigeon-hole principle.
Firstly, let us use congruence classes modulo n, as it simplifies the situation. Now the question is:
From any "assortment" of $2n-1$ numbers in $[0,1,2,3...(n-1)]$ is there a "sub-assortment" of n numbers $[a_{0},a_{1},a_{2},...a_{n}]$ such that $a_{0}+a_{1}... a_{n} \equiv 0$ mod n
Now, to see some examples of such a "sub-assortment":
There are 2 possible assortments of 2 numbers which add up to a multiple of 2: $[0,0],[1,1]$
There are 4 possible assortments of 3 numbers which add up to a multiple of 3: $[0,0,0],[1,1,1],[2,2,2],[0,1,2]$
There are 8 possible assortments of 4 numbers which add up to a multiple of 4: $[0,0,0,0],[1,1,1,1],[2,2,2,2],[3,3,3,3],[0,0,2,2],[0,0,1,3],[0,2,2,3],[0,1,1,2]$
It seems as if there are always $2^{n-1}$ assortments of $n$ integers which add up to a multiple of $n$ (Though I cannot seem to prove it.)
Using the "stars and bars" approach we know there are $2n-1\choose{n}$ different ways of selecting ANY assortment of $n$ integers mod n. (If you have not heard of this, look up the numberphile youtube video of the same name.)
This is because the "bars" represent the dividers between congruence classes
Thus, if there was a way to choose $2n-1$ integers so than NO $n$ integers have the property (of their some being $\equiv 0$) there could not be more than $2n-1\choose n$ $-2^{n-1}$ different assortments of n integers to select, or else on of them would have to have the property.
Furthermore, we know any "non-trivial" assortment of $2n-1$ integers mod n must contain at-least 3 different congruences in it, or there would be $n$ of one conguence which would form a trivial solution.
Now, I cannot drive home the solution. I do not know if I am tantalizingly close or far off. Can someone help by either:
- Proving the claim made above
- Providing a different approach
- Giving me a hint on how to finish this approach
