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Aug 13, 2020 at 17:26 vote accept BPP
Aug 13, 2020 at 17:06 comment added Max Alekseyev @GerhardPaseman: We should have $j_i$ as divisors of $n$ as otherwise "negative" differences would not be multiples of $j_i$. Ideally $n$ should be LCM of the $j_i$. For $k=3$ and $n=200$, we have with necessity that $m\leq 34$. See my answer for details.
Aug 13, 2020 at 14:33 answer added Max Alekseyev timeline score: 1
Aug 13, 2020 at 5:11 comment added BPP I think the sets work for $n\ge m$. I've edited my question to fix this. Thanks for pointing it out.
Aug 13, 2020 at 5:10 history edited BPP CC BY-SA 4.0
fix wrong inequality
Aug 13, 2020 at 4:22 comment added Gerhard Paseman Have you tried k different coprime integers j_i?. Then for N less than the product of the j_i, the sets which are 0 mod j_i and have about N/j_i elements each should work. For N=200, the numbers 5,6, and 7 do a nice job. I'm not sure you will do much better than m=200/7 for k=3 and N=200. Gerhard "Residue Classes Can Be Shifty" Paseman, 2020.08.12.
Aug 13, 2020 at 4:18 comment added Max Alekseyev It is necessary and sufficient that for any nonzero $d\in\Bbb Z/n\Bbb Z$, there exists $i\in\{1,2,\dots,k\}$ such that $d\notin (A_i-A_i)$.
Aug 13, 2020 at 3:58 history edited Max Alekseyev CC BY-SA 4.0
m was mixed with k
Aug 12, 2020 at 23:04 history edited BPP CC BY-SA 4.0
added 5 characters in body
S Aug 12, 2020 at 21:11 history edited BPP CC BY-SA 4.0
Math Jaxed
S Aug 12, 2020 at 21:11 history suggested Daniele Tampieri CC BY-SA 4.0
Math Jaxed
Aug 12, 2020 at 21:05 review Suggested edits
S Aug 12, 2020 at 21:11
Aug 12, 2020 at 20:26 review First posts
Aug 12, 2020 at 21:43
Aug 12, 2020 at 20:24 history asked BPP CC BY-SA 4.0