Discuss the following. I got a good platform to remove all me quarries from my mind by positing the problems like this. Thanks again for support.
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 $(\mathbb Z_M, +)$.
On
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.
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 $\lbrace 2,3,7,8,12,13 \rbrace$.