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How long a series of points in (0,1) can be chosen such that the first two are in different halves, the first three are in different thirds, ... the first $n$ are in different $n^{\text{th}}$s? My first try of $(0+,1-,\frac{1}{2}-,\frac{3}{4}-, \frac{1}{5}+,\frac{5}{8}-,\frac{1}{3}-,\frac{7}{8}-,\frac{1}{3}+)$ works through $9$, but there are two points in $(0.3,0.4)$. The plus and minus signs indicate a shift of some distance away from the given point small enough not to move over any fraction of interest.

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Mathworld calls this the "18-point problem": – joriki Jun 5 '11 at 2:06
up vote 13 down vote accepted

An answer is asserted but not proved at the following URL:

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That is very surprising. I'm glad I didn't spend more than a few minutes on this! I never would have reached anything close to 17. – Fahad Sperinck Jun 5 '11 at 1:59
Fascinating. Interestingly, the solution roughly matches Ross' up to $5/8-$. – joriki Jun 5 '11 at 1:59
See also… and the freely available original proof at… – joriki Jun 5 '11 at 2:04

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