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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 $(\mathbb Z_M, +)$.

Batominovski
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GAn
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2 Answers2

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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.

zyx
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    @YACP: I have an answer once the OP clarifies the question. There is no point in writing out details until then. – zyx Mar 17 '13 at 09:40
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    I stated what was unclear. If @YACP can explain what is being selected that would be progress. (set or multiset) – zyx Mar 17 '13 at 09:43
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    There is a counterexample in that case, such as taking M > 3 and selecting any number of elements equal to 1. I suspect the OP is looking for solutions of $x + y = 2z$, or some other interpretation in which the problem has substance. But it is not worth my time to invent every possible useful interpretation and solve it. It is enough to state that E-Z-G is the relevant universe of ideas and that some clarification is in order. Have a nice day. @YACP – zyx Mar 17 '13 at 09:47
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    To the idea that there is a minimum number of elements forcing a solution of $x+y+z=0$. The first question asks to find that minimum. Did you not understand that? @YACP – zyx Mar 17 '13 at 09:51
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    I deleted the comment noting a particular user's voting history. The first part referred to a comment now deleted, and the second part was off-topic. It also seems that the rest of the comments also refer to deleted comments. – robjohn Mar 17 '13 at 18:11
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    @robjohn, the best thing would be to restore all moderator-deleted comments during the ongoing discussion on meta, but if you don't intend to do that, at least please remove your comment (and then this one). The information about comments being deleted has been conveyed to me, to YACP, and anyone else who is interested enough to read meta. Leaving it in place, as you can see from the upvotes, ends up as a solicitation of votes to "ratify moderator decision", when nobody can check whether the content is as claimed, or whether the deletion was necessary. – zyx Mar 18 '13 at 05:27
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    Just let us know when you are done with this work-in-progress flurry of comment deletions, involuntary moderator deletions, flags, meta questions, revised meta questions, closed meta questions, downvotes, (suspensions), or whatever. When the final version is baked enough to be ready for prime time, I can consider whether to reply. Until then it is an odd sort of theatric performance. – zyx Mar 18 '13 at 09:13
  • The inbox gets spammed every time there is another 5 comments, of which 3 are deleted, then the embarrassing parts edited out of the remaining ones (repeat as necessary). This is the kind of thing you don't have to follow, it follows you. (EDIT: the comment this answers just disappeared!! Something about me "following" someone's posts. Great stuff, but tiring.) – zyx Mar 18 '13 at 09:25
  • @zyx: deleted comments cannot be restored (except possibly by a dev). I left a comment explaining why I deleted a comment rather than simply deleting (and then replying to another comment as to why I deleted it). It was a courtesy; I was not looking for ratification. The idea of moving a discussion to meta is so that it can be deleted to clean up the Q&A site. In any case, inappropriate comments will be deleted, meta discussions notwithstanding. If needed and appropriate, a moderator can quote a deleted comment on meta. – robjohn Mar 18 '13 at 13:20
  • @robjohn, thanks, I didn't know deletions are irreversible. It was understood that the explanation was a courtesy, but that is fully consistent with (and was assumed when making) my request. As far as this answer goes, I'll either move the mathematical content of the comments into the answer, or leave it as is while the meta discussion is in process and delete later. – zyx Mar 18 '13 at 15:50
  • @zyx: This comment and the cited MSO question and answers might be enlightening here. – robjohn Mar 18 '13 at 16:26
  • @robjohn, in line with the earlier request, I really think the moderators should be careful with words like "inappropriate comments will be [and by implication, were] deleted". This is characterizing comments, and the actions or judgement of the comment authors, when the evidence has been destroyed. It is more accurate to say "comments that the moderators consider inappropriate will be deleted" which acknowledges the possibility that others might consider things differently. – zyx Mar 19 '13 at 01:38
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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 $\lbrace 2,3,7,8,12,13 \rbrace$.

Ewan Delanoy
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