Combinatorics in finite cyclic groups Discuss the following. 
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1) Find the minimum elements must be selected from the group $(\mathbb Z_M, +)$, where $M = 2k$ such that among the selected elements surely there exist three (not necessarily distinct) with sum $0$.
2) Discuss the same in case of $M = 15$ and $(\mathbb Z_M, +)$.
 A: This is very similar to the Erdos-Ziv-Ginzburg theorem.
http://en.wikipedia.org/wiki/Zero-sum_problem
The answer depends on whether you are selecting a subset or a multiset from the group.
A: Question 1 The minimum number you are looking for is exactly $k+1$. The proof below also shows that there is a unique optimal counterexample (the set of odd elements).
 Lower bound  The set of odd elements in ${\mathbb Z}_M$ has cardinality
$k$, and a sum of three odd numbers is odd, and therefore nonzero.
 Upper bound  Let $A \subseteq {\mathbb Z}_M$ have the property that no
sum of three (not necessarily distinct) elements of $A$ is zero. Suppose that $A$ contains an even number, $2j$ for some $j\in {\mathbb Z}_M$. Then $-j\not\in A$ and $k-j\not\in A$, for otherwise we could write $0=x+x+2j$, where $x$ is $-j$ or $k-j$. Let $Y={\mathbb Z}_M \setminus \lbrace -j,k-j \rbrace$ ; we thus have $A\subseteq Y$. Define the map
$i : Y \to Y$  by $i(t)=-2j-t$. Then $i(t) \neq t$ for any $t\in Y$, and $A$ cannot contain both $t$ and $i(t)$,  for otherwise we could write
$0=t+i(t)+2j$. So $A$ contains at most half the elements of $Y$, and hence $|A| \leq k-1$.
If, on the other hand, $A$ does not contain any even element, then $A$ consists only of odd elements, so $|A| \leq k$ and we are done.
 Question 2 Arguing similarly, for $M=15$ one sees that the
minimal number is $7$, and that there are two optimal counterexamples :
$\lbrace 1,4,6,9,11,14 \rbrace$ and $\lbrace 2,3,7,8,12,13 \rbrace$.
