Jump to content


Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.

Of course, you can also enjoy our collection of amazing optical illusions and cool math games.

If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.

Thanks and enjoy the Den :-)
Guest Message by DevFuse
 

Photo
* * * * * 2 votes

How many heats?


  • Please log in to reply
19 replies to this topic

#11 unreality

unreality

    Senior Member

  • Members
  • PipPipPipPip
  • 6370 posts

Posted 28 February 2008 - 11:40 PM

Spoiler for my solution


No, yours doesn't work, my friend :P The fastest of each heat might not necessarily be the 5 fastest because the heats are random. Two of the five fastest could have run in the same initial heat. My solution is the best one so far, from what I can tell. Bonanova never did give his answer
  • 0

#12 frog101858

frog101858

    Newbie

  • Members
  • Pip
  • 2 posts

Posted 17 March 2008 - 10:50 AM

Okay, call me slow but I am just not getting this.. The numbers in red you can rule out due to at least 3 other horses being faster..

1: A1 A2 A3 A4 A5
2: B1 B2 B3 B4 B5
3: C1 C2 C3 C4 C5
4: D1 D2 D3 D4 D5
5: E1 E2 E3 E4 E5

6: A1 B1 C1 D1 E1 Winner of this heat is the fastest #1 (C1 and D1 eliminated due to 3 faster scores, based on arbitrary times)

So this is where I'm not following... Who's to say that E2 isn't faster than B1

Just to try to prove my point on this, I arbitrarily assigned times to these 15 horses

A1 (2) A2 (6) A3 (7)
B1 (6) B2 (7) B3 (9)
C1 (8) C2 (9) C3 (10)
D1 (10) D2 (11) D3 (12)
E1 (3) E2 (4) E3 (5)

And if C1 and D1 were the fastest of the horses in heat 3 and 4 then that means all other horses in their heats were slower than A1 B1 or E1
If B1 was 3rd fastest in this heat then B2 and B3 must be slower

So now we have 6 horses left
A2 A3
B1

E1 E2 E3


This is where I get lost... I don't understand this part: "[color="#FFFF00"]So, send home the horse who got third place against the second place horses from this race"

So if we did that we would take out E3

.......................................................

Alright it finally hit.. and even though I was tempted to delete this now since I figured it out, I thought I'd go ahead and post it anyways for other people who are as slow as I am... :0)


I was thinking that according to my numbers if we took out E3 that it was a faster time than the others remaining, and I couldn't figure out what justified taking it out. But if it came in 3rd place in heat 5 and E3 came in 2nd place in heat 6, then that means that E1, E2 and A1 are all faster, therefore justifying ruling this one out... Eureka!!!

A2 A3
B1

E1 E2


And now there are only 5 horses to race in HEAT 7

oh my.. only took me 2 hours.. Don't I feel stupid!!
  • 0

#13 brhan

brhan

    Advanced Member

  • Members
  • PipPipPip
  • 466 posts

Posted 17 March 2008 - 11:49 AM

No, yours doesn't work, my friend :P The fastest of each heat might not necessarily be the 5 fastest because the heats are random. Two of the five fastest could have run in the same initial heat. My solution is the best one so far, from what I can tell. Bonanova never did give his answer


... and your solution was ...

* Split them into five heats. Take the 3 fastest in each heat, meaning you now have the 15 fastest.
* So do 3 heats now. Take the 3 fastest in each heat. You have the 9 fastest horses.
* Do a single heat with 5 random horses of the 9, cutting out the 2 last-place horses, leaving you with 7.
* Take another random 5 horses and do it again, cuttnig out 2, leaving you with 5 horses.
* Do one final heat and take the top 3 from that race.

Those top 3 horses will only have to race 5 times each.


It works ... but a lot races ....
The top 3 horses need to run 3 times, and in total seven races. If you don't understand my approach, there are other posts with the same solution but different approach. You may need to optimize your procedure ...

Edited by brhan, 17 March 2008 - 11:50 AM.

  • 0

#14 Duh Puck

Duh Puck

    Advanced Member

  • Members
  • PipPipPip
  • 445 posts

Posted 17 March 2008 - 03:38 PM

Nice problem. I was stuck on eight heats too for a bit, just like frog101858, for basically the same reason. Once I popped into Excel and started charting it out, it clicked. Seven heats indeed.
  • 0

#15 NintendoMad

NintendoMad

    Junior Member

  • Members
  • PipPip
  • 66 posts

Posted 01 November 2008 - 11:24 PM

Damn I thought it was 6, 9 or 11!!! AHHHHH!! I get it from the spoiler!!! 7!! I know why know!! I understand from PDR and roolstar!!
  • 0

#16 tygar

tygar

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 13 November 2008 - 06:59 AM

question?? y cant u just do 6 heats?? i havent read all the posts but it seams logical to me that u divide the 25 into 5 heats the 1st place winners of the 5 heats makes 5 horses that race again the fastest is duh the fastest the second is duh the second fastest and the third is duh the third fastest just 6 heats needed seems logical to me
  • 0

#17 bonanova

bonanova

    bonanova

  • Moderator
  • PipPipPipPip
  • 5824 posts
  • Gender:Male
  • Location:New York

Posted 13 November 2008 - 09:26 AM

question?? y cant u just do 6 heats?? i havent read all the posts but it seams logical to me that u divide the 25 into 5 heats the 1st place winners of the 5 heats makes 5 horses that race again the fastest is duh the fastest the second is duh the second fastest and the third is duh the third fastest just 6 heats needed seems logical to me

Hi tygar, and welcome to the Den.

What if the 5 fastest horses were in the first heat?
  • 0
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#18 NAPALM71

NAPALM71

    Newbie

  • Members
  • Pip
  • 4 posts
  • Gender:Male
  • Location:CA

Posted 31 August 2012 - 01:33 AM

Spoiler for my solution


ummm this is not a solution it is an answer in which you gave no explanation meaning it is NOT a solution
  • 0

#19 mmiguel

mmiguel

    Advanced Member

  • Members
  • PipPipPip
  • 134 posts
  • Gender:Not Telling

Posted 02 September 2012 - 08:37 AM

The annual Horse racing tournament is a week away.
You own 25 horses, and you're allowed to enter up to three.
You decide to hold a series of races to find your three fastest horses.
You have access to a track that permits five horses to race against each other.

Here's the deal:

[1] No two of your horses are equally fast - there will be no ties.
[2] You have no stopwatch - you can't compare performances from different heats.
[3] You do not want to tire the horses needlessly - the fewer heats needed, the better.
[4] You want to know which horse is fastest, which is next fastest and which is third-fastest of the 25.

How many heats are needed?

Spoiler for try

  • 0

#20 mmiguel

mmiguel

    Advanced Member

  • Members
  • PipPipPip
  • 134 posts
  • Gender:Not Telling

Posted 02 September 2012 - 08:46 AM

* Split them into five heats. Take the 3 fastest in each heat, meaning you now have the 15 fastest.
* So do 3 heats now. Take the 3 fastest in each heat. You have the 9 fastest horses.
* Do a single heat with 5 random horses of the 9, cutting out the 2 last-place horses, leaving you with 7.
* Take another random 5 horses and do it again, cuttnig out 2, leaving you with 5 horses.
* Do one final heat and take the top 3 from that race.

Those top 3 horses will only have to race 5 times each.


In my solution, the top horse races twice, the 2nd and 3rd to best horses race 3 times each (the max for any horse).
  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users