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Dec 20, 2020 at 9:11 history edited katago CC BY-SA 4.0
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Dec 20, 2020 at 9:07 comment added katago Zhi-Wei Sun: thanks, I will check this.
Dec 20, 2020 at 9:03 history edited katago CC BY-SA 4.0
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Dec 20, 2020 at 8:01 comment added Zhi-Wei Sun For any exact $m$-cover $A=\{a_s+n_s\mathbb Z\}_{s=1}^k$, I proved in a 1992 paper [Israel J. Math.] that for each $n=1,\ldots,m$ we have $\sum_{s\in I}\frac1{n_s}=n$ for some $I\subseteq\{1,\ldots,k\}$.
Dec 20, 2020 at 4:55 history edited katago CC BY-SA 4.0
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Dec 20, 2020 at 4:50 history edited katago CC BY-SA 4.0
added 260 characters in body
Dec 20, 2020 at 4:43 history edited katago CC BY-SA 4.0
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Dec 20, 2020 at 4:35 history answered katago CC BY-SA 4.0