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

- - - - -

Practical number problem

  • Please log in to reply
3 replies to this topic

#1 Trackcoach



  • Members
  • Pip
  • 2 posts

Posted 09 January 2014 - 06:51 PM

Hello all!  I have a practical problem that has really been bugging me.  I do not have the answer.  Maybe you can help!


I need some help making up a track and field schedule for our athletic conference (SWC). 


10 teams in the conference

5 weeks of competition (I don't think it is possible with 4 weeks of competition)
1 competition a week, each is either a 2 or 3 team competition(dual or tri)
All teams see each other 
No teams see each other twice
Host school rotates so schools have a mix of home and away meets
A school can have one bye week if needed
Four years of numbers needed

See the attachment for a 7-8 team schedule.  I cannot figure it out for 10 teams.


Thanks and best of luck!

Attached Files

  • 0

#2 k-man


    Senior Member

  • Members
  • PipPipPipPip
  • 537 posts
  • Gender:Male

Posted 10 January 2014 - 05:01 PM

Welcome to the forum, Trackcoach.


It's impossible to make a schedule that would satisfy ALL of your requirements. Something needs to change. For example, you should be able to do it in 6 weeks, but not in 5. Here is why...


Given that there are 10 teams and every team must meet every other team exactly once there should be a total of 45 head-to-head events. To fit it in 5 weeks you would need to hold 9 events every week. The only way to do that is to have 3 teams compete in one location (one host and 2 guests) and do it in 3 locations. This will require one team to get a buy every week. So far, so good. The problem is that the team that gets a buy will only compete in 4 remaining weeks and can therefore see only 8 other teams, but it needs to see 9. If you allow 4 teams to meet in one location (albeit not all 4 get to compete with all others at that time) then you should be able to do it in 5 weeks.


Hope this helps.

  • 0

#3 Trackcoach



  • Members
  • Pip
  • 2 posts

Posted 10 January 2014 - 09:18 PM

It does, thanks, I appreciate your response.  If the teams saw 8 other teams instead of 9, that would be ok, but we cannot put four teams together, it would be too many kids.


Thanks again for your insight.  I'll go back to the drawing board and try to come up with a schedule that fits these perameters.

  • 0

#4 dgreening


    Junior Member

  • Members
  • PipPip
  • 98 posts
  • Gender:Male
  • Location:Maryland [DC area]

Posted 17 January 2014 - 07:35 PM

K-Man is right, you cannot organize in a way to have each team play 9 teams in 5 weeks


But with different constraints, there is an answer.


8 team in 5 weeks

You can set up 4 tournaments/ week and have each team play 8 of the other teams


at 2 schools, 3 teams will compete

  • School A hosts B&C
  • D hosts E&F

at 2 schools 2 teams compete

  • G hosts H; and
  • I hosts J

Over 5 weeks each school will play 3 times in a 3 team event and twice in a 2 team event


At the end of 5 weeks each school will have played 8 teams [2 teams x 3 weeks + 1 team x 2 weeks].


9 teams in 6 weeks


If you want to have everyone play each other team, you must add a 6th week. in which you will have five 2-team events

  • 0

0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users