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Written by Kuram

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Written by CMaster


Nope, what if the top three hourses are in the same group. Taking one leader from each group is wrong.


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Written by didxga

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Written by egonjensen


Seven.
The first 5 to find the 3 fastest in each group and the sixth with the winners of each group.
Now we know the fastest horse.
The second fastest horse must either be the second fastest in the sixth race or the second fastest horse in the final winners first race.
In the first case, the third fastest horse can either be the third fastest horse in the sixth race or the second fastest horse in either the fastest or second fastest horses first race.
Otherwise the third fastest horse could either be the second fastest horse in the sixth race or the third fastest horse in the final winners first race.
This makes for a total of five horses for the seventh, and final, race.


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Written by you


It requires 8 Iterations in total: (divide horses in to 5 groups: A,B,C,D,E)
Race(1 - 5): Race 5 horses each group and let Winner horses are A1,B1,C1,D1,E1.
Race-6: Race A1,B1,C1,D1,E1. This will give the fastest horse. Let say C1 is the fastest horse.
Race-7: The second fastest could be the second fastest in C group (C2) or fastest of other groups. So race A1,B1,C2,D1,E1. Let say B1 won. So B1 is the second fastest of all.
Race-8: The third fastest could be the second fastest of in B group (B2) as B1 is the over all second fastest horse. So race A1,B2,C2,D1,E1.
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