KGS math club/hints 11 5

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Try to start with a small solution (e.g. three horses or two horses), and show how to build up solutions with any number of horses (including 9).

Suppose two horses have integer speeds $y$ and $x$ with $y>x$ (think of them as laps per day, or whatever you like, the name of the time period doesn't matter). Assume the horses start together at the wide section of track, then the faster horse will do exactly $y$ laps of the course in the day, and the slower does exactly $x$ laps -- so it does $y-x$ extra laps. This means the faster horse has met/overtaken the slower horse exactly $y-x$ times (including this final meeting at the end of the day). These meetings will be equally spaced over the day (by symmetry -- for example by running time backwards). These $n$ meeting times will be: \[\frac{1}{y-x}, \frac{2}{y-x}, \ldots, \frac{n}{y-x}=1;\] so for these meetings to match up with the times the faster horse is back at the wide section of track we need these times to be inside the list $\frac{1}{y}, \frac{2}{y}, \ldots, \frac{y}{y}=1$, i.e. $y-x$ needs to divide $y$ perfectly. This condition is also enough to guarantee the two horse only ever meet back at the start. We need this property to hold separately for every pair of horses in the race.

So now the problem has been reduced to one of trying to find lists of distinct speeds with the property that when you take any pair the difference between then divides the bigger one.