2 Really Hard Questions

[hobbes]

deleted because it was a flawed number.

[Chevy]

How did you get 9 (apart from Yonki's solution)?


I got as far as 10:

Round 1: 25 runners, 5 races=the top 2 in each going through to round 3. The 5 third place contestants go into round 2. <5 races>

Round 2: 5 runners (3rd placers in Round 1) , one race = the fastest one going into round 4 <1 race>

Round 3: 10 runners (top 2 from each race in round 1), 2 races, the top 3 in each go to round 4 <2 races>

Round 4: 7 runners (6 from round 3, 1 from round 2), 2 given a bye, one race of the remaining 5 the top 3 going into round 5. <1 race>

Round 5: 5 runners (top 3 from round 4 and the 2 from round 4 with the bye) Top 3 are the overall top 3. <1 race>


(sorry if it&#39;s not that clear)

[yonki]

ok sorry,7 is the right number

same as before 5 races and pick the first 3 of each round.now a race between the 5 winners.now you have the fastest.and 6 races.you still have to race the 2nd and 3rd of the first race and the 2nd and 3rd of th all winners race.am i wrong again?

[Chevy]

Originally posted by yonki@24 April 2004 - 22:41
ok sorry,8 is the right number

Unless you explain how you got it I think you are playing "guess the number"

[Chevy]

Originally posted by yonki@24 April 2004 - 22:41
ok sorry,7 is the right number

same as before 5 races and pick the first 3 of each round.now a race between the 5 winners.now you have the fastest.and 6 races.you still have to race the 2nd and 3rd of the first race and the 2nd and 3rd of th all winners race.am i wrong again?

The 5 fastest arn&#39;t neccessarily the winners from each of the first 5 races - they could all be in the first race.

Even if they were, it would also result in the 6th fastest person being the winner in you "2nd placers race"

[DanB]

I think we need a maths world

[yonki]

Originally posted by Chevy+24 April 2004 - 23:44--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Chevy @ 24 April 2004 - 23:44)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-yonki@24 April 2004 - 22:41
ok sorry,8 is the right number

Unless you explain how you got it I think you are playing "guess the number" [/b][/quote]
mekivoke,i meant 7

[muchspl2]

7 is correct

First Groups:
1.2.3.4.5 | 6.7.8.9.10 | 11.12.13.14.15 | 16.17.18.19.20 | 21.22.23.24.25

4,5,9,10,14,15,19,20,24,25 get eliminated since you know for a fact that there are atleast 3 racers faster than them, and we are looking for the top 3.

Remaining groups: 1.2.3 | 6.7.8 | 11.12.13 | 16.17.18 | 21.22.23

Racing the fastest of these groups (1,6,11,16,21) allows us to eliminate all racers in the groups with the slowest of the fastest (17,18,22,23) We can also eliminate 12 and 13 since we know theres atleast 3 faster racers 1, 6 and 11. We can eliminate 8 also because we know that 8 is slower than 7 and 6, and 6 is slower than 1. This makes there atleast 3 faster runners than 8.

left over now are: 1.6.11.2.3.7

We know for sure that 1 is the fastest, so we can race the remaining 5 racers and figure out the second and third fastest.

The races look like this

Race 1: 1.2.3.4.5 Results: Eliminate 4 and 5 (1,2,3 are faster)
Race 2: 6.7.8.9.10 Results: Eliminate 9 and 10 (6,7,8 are faster)
Race 3: 11.12.13.14.15 Results: Eliminate 14 and 15 (13,14,15 are faster)
Race 4: 16.17.18.19.20 Results: Eliminate 19 and 20 (18,19,20 are faster)
Race 5: 21.22.23.24.25 Results: Eliminate 24 and 25 (21,22,23 are faster)
Race 6: 1.6.11.16.21 Results: Eliminate 16,21,12,13,17,18,22,23,8 (1,6,11 are faster)
Race 7: 2.3.6.7.11 Results: Eliminate 6,7,11 (1,2,3 are faster)

The reasoning behind every elimination is that the race proves there&#39;s atleast 3 faster runners than those eliminated.


RESULT: 7 Races.

[Chevy]

Ohhhhhhhhh



Thanks Muchspl, was fun (took longer to work out the given answer than it did to try and guess before though )

[J'Pol]

Good novel thread, bit of a change from the usual stuff.

Well done that man, thanks for that.