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Aug 28, 2014 at 12:19 vote accept Konstantinos Gaitanas
Sep 15, 2013 at 21:53 answer added Douglas Zare timeline score: 1
Sep 13, 2013 at 17:07 comment added Konstantinos Gaitanas @DouglasZare you are right, i took a quick look at the abstract while i was too tired to focus at everything the article says and i supposed that the answer was there but i'd better read it later.It has some connection with my question but it doesn't seem to answer it- at least not with a little effort and some mathematical extensions from my side.
Sep 13, 2013 at 15:09 comment added Douglas Zare While the paper by Odlyzko and Stanley is related and interesting, it doesn't seem to answer your question since it looks at unrestricted subsets, so why accept that as the answer? Are you no longer interested in establishing results about subsets of a given cardinality?
Sep 13, 2013 at 14:24 vote accept Konstantinos Gaitanas
Sep 13, 2013 at 17:05
Sep 13, 2013 at 14:24 vote accept Konstantinos Gaitanas
Sep 13, 2013 at 14:24
Sep 12, 2013 at 15:00 comment added The Masked Avenger For large n, take the smallest k-1 elements of a putative k subset of n. Find an element c larger than any of those but smaller than n, such that the sum will be 0 mod m. Most of the time, about 1/m out of the available choices for c will work. There are ways to make this more rigorous and more precise, but that is the basic idea.
Sep 12, 2013 at 8:21 comment added Konstantinos Gaitanas I made some edits needed as Ira Gessel noted.I am sorry for maybe confusing you,i want the sum to be a multiple of m. Is there a proof for the asymptotic value 1/k? Asymptotic estimates would do the job for me by the way!
Sep 12, 2013 at 8:14 history edited Konstantinos Gaitanas CC BY-SA 3.0
deleted 5 characters in body
Sep 12, 2013 at 3:17 answer added Ira Gessel timeline score: 3
Sep 12, 2013 at 0:45 answer added Michael timeline score: 1
Sep 11, 2013 at 23:22 comment added The Masked Avenger Asymptotically (when n much larger than m or k) about 1/m of them. How precise a result do you need?
Sep 11, 2013 at 22:14 history asked Konstantinos Gaitanas CC BY-SA 3.0