Wikipedia:Reference desk/Archives/Mathematics/2018 April 13

= April 13 =

Largest multiset that cannot be partitioned in two equal sum sets
Inspired by the above "addition game". Let M be a fixed integer. For which N, if any, is it guaranteed that a set of N integers (not necessarily distinct) all between 1 and M can be partitioned into two subsets with the same sum? What about the same sum modulo M? (Or, conversely, assume that you are given M and N, can you exhibit N integers such that no partition of the set in two subsets produces the same sum?)

Formally: given M, find (if possible) the smallest N such that

$$\forall a_1, a_2, ... a_N \in \{1,2,...M\}, \exist b_1, b_2, ... b_N \in \{-1,1\}, \sum_1^N a_i b_i = 0 \text{ (maybe } mod\ M \text{)}$$

I would assume that there is a maximum N in both cases, but cannot prove it in either. The second case (modulo M) looks like it ought to be solved with a smart pigeonhole principle but I did not find it. Tigraan Click here to contact me 14:54, 13 April 2018 (UTC)
 * I'm probably missing something, but why can't you just take N integers whose sum is odd? You'd then be guaranteed not to be able to divide it into parts with equal sums. --RDBury (talk) 00:16, 14 April 2018 (UTC)
 * Huh... yeah, of course. (For the "modulo M" case, it might not work if M is odd, but a similar idea applies: take numbers all but one equal to 0 mod M.) Sounds trivial in retrospect. Tigraan Click here to contact me 20:30, 15 April 2018 (UTC)