Guest Posted December 20, 2008 Report Share Posted December 20, 2008 (edited) Eleven theater groups took part in a festival. Every day some of the groups put on their plays while the others watched. When the festival was over it was possible to affirm that each one of the groups had been able to attend, at least once, the performances of each of the other groups. What is the minimum number of days that the festival lasted? Good luck and have fun!-zak Edited December 20, 2008 by Riddle Master Zack Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 20, 2008 Report Share Posted December 20, 2008 22 days. Eleven groups each watched ten performances = 110 "watchings". Each day, no more than five groups can watch a performance. 110/5 = 22 is therefore a lower bound. It make take more than 22 days - still thinking. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 20, 2008 Report Share Posted December 20, 2008 Have I missed something? I think I've missed something because here is my answer: Only one group performs each day while each other group watches. Eleven groups = eleven days. I think this number might be able to be reduced if we assume that groups perform multiple times during the day, allowing speculating groups to see more than one during the day. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 20, 2008 Report Share Posted December 20, 2008 5 of the groups perform in one day while the other 6 watch and the next day the other 6 groups perform while the 5 groups who had already performed watched. So I'm Thinking 2 Days? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 20, 2008 Report Share Posted December 20, 2008 You could always say 1 day . depending on how long the play was I mean if it was an hour play all groups could have performed while all the others watched when it wasn't thier turn Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 21, 2008 Report Share Posted December 21, 2008 I feel that i must have missed something, but here's my guess... 2 days. On the first day, every one but one group (group #1) performs. Hence, all the groups have seen every play but #1's and their own. On the second day, only number one needs to perform with everyone else watching, and VOILA, all groups have seen all plays... or have i blown all semblence of logic? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 21, 2008 Report Share Posted December 21, 2008 (edited) 22 days. Eleven groups each watched ten performances = 110 "watchings". Each day, no more than five groups can watch a performance. 110/5 = 22 is therefore a lower bound. It make take more than 22 days - still thinking. You're thinking too hard. Try to Combine more of the days. (no one right yet) Edited December 21, 2008 by Riddle Master Zack Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 21, 2008 Report Share Posted December 21, 2008 use numbers like 1 2 3... for each group (each day more than 1 group can perform)(and one group can perform more than a day (day x and y) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 21, 2008 Report Share Posted December 21, 2008 5 of the groups perform in one day while the other 6 watch and the next day the other 6 groups perform while the 5 groups who had already performed watched. So I'm Thinking 2 Days? I feel that i must have missed something, but here's my guess... 2 days. On the first day, every one but one group (group #1) performs. Hence, all the groups have seen every play but #1's and their own. On the second day, only number one needs to perform with everyone else watching, and VOILA, all groups have seen all plays... or have i blown all semblence of logic? There's a problem with your logic. The groups won't have seen the performances of the other groups which perform on the same day. Well, are we in an agreement that it is possible in eleven days? Only one groups performs per day. If that is not the answer, then it must be less. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 22, 2008 Report Share Posted December 22, 2008 Sorry about the double post. I was going to just sit round and be lazy as I figured it would be too hard to solve. But I changed my mind. Here's the lowest answer I've gotten so far: 7 Days (1 week) To get this, I drew and 11x11 grid with the length representing the viewing groups while the height represented the performing groups. I then proceeded to try and take the biggest chunks of the grid out every day. I also figure that if there were twelve groups, they could still get through in seven days as well. Day 1: 1, 2, 3, 4, 5 are watched by 6, 7, 8, 9, 10, 11 Day 2: 6, 7, 8, 9, 10, 11 are watched by 1, 2, 3, 4, 5 Day 3: 1, 2 are watched by 3, 4, 5 6, 7, 8 are watched by 9, 10, 11 Day 4: 3, 4, 5 are watched by 1, 2 9, 10, 11 are watched by 6, 7, 8 Day 5: 1 is watched by 2 3 is watched by 4, 5 6 is watched by 7, 8 9 is watched by 10, 11 Day 6: 2 is watched by 1 4 is watched by 3, 5 7 is watched by 6, 8 10 is watched by 9, 11 Day 7: 5 is watched by 3, 4 8 is watched by 6, 7 11 is watched by 9, 10 This is the best I can come up with. In the later days, groups would be able to watch groups they've already seen of course. Eg, by day 7, groups 1 and 2 have seen every performance and been seen by every performance so they may as well go home. This answer doesn't feel quite as efficient as it could be so I figure someone might be able to find an answer for 6 days but that would be as low as it would go. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 22, 2008 Report Share Posted December 22, 2008 I got 6 days, however I don't have time to write it out as much as you did. It might be 5. Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted December 22, 2008 Report Share Posted December 22, 2008 I got 6 days, however I don't have time to write it out as much as you did. It might be 5. Six seems to be the minimum Group1 performs on days 1 and 2 Group2 performs on days 1 and 3 Group3 performs on days 1 and 4 Group4 performs on days 1 and 5 Group5 performs on days 2 and 3 Group6 performs on days 2 and 4 Group7 performs on days 2 and 5 Group8 performs on days 3 and 4 Group9 performs on days 3 and 5 Group10 performs on days 4 and 5 Group11 performs on day 6 All groups are able to preform for all the others in six days. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 22, 2008 Report Share Posted December 22, 2008 Eleven theater groups took part in a festival. Every day some of the groups put on their plays while the others watched. When the festival was over it was possible to affirm that each one of the groups had been able to attend, at least once, the performances of each of the other groups. What is the minimum number of days that the festival lasted? Good luck and have fun!-zak Questions arise:On a given day, are all performances given simultaneously?Can a group perform more than once in a day?On a given day, can a group both perform and at another time watch the performance of a different group?Normal interpretation of the OP seems to rule out question 3, but the answers to questions 1 and 2 certainly affect the answer. To clarify regarding question 1. If Group A and Group B perform on a particular day, can Group C watch both performances? Not if they are simultaneous. To clarify regarding question 2. Could Group A perform 10 times in one day so all the other Groups could be sure to see them that day? Not if a group can perform only once per day. To clarify regarding question 3. Could all the Groups see all the other groups perform in a single day? Not if a group can only perform OR watch in a single day, or can only perform once in a day. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 22, 2008 Report Share Posted December 22, 2008 Sorry about the double post. I was going to just sit round and be lazy as I figured it would be too hard to solve. But I changed my mind. Here's the lowest answer I've gotten so far: 7 Days (1 week) To get this, I drew and 11x11 grid with the length representing the viewing groups while the height represented the performing groups. I then proceeded to try and take the biggest chunks of the grid out every day. I also figure that if there were twelve groups, they could still get through in seven days as well. Day 1: 1, 2, 3, 4, 5 are watched by 6, 7, 8, 9, 10, 11 Day 2: 6, 7, 8, 9, 10, 11 are watched by 1, 2, 3, 4, 5 Day 3: 1, 2 are watched by 3, 4, 5 6, 7, 8 are watched by 9, 10, 11 Day 4: 3, 4, 5 are watched by 1, 2 9, 10, 11 are watched by 6, 7, 8 Day 5: 1 is watched by 2 3 is watched by 4, 5 6 is watched by 7, 8 9 is watched by 10, 11 Day 6: 2 is watched by 1 4 is watched by 3, 5 7 is watched by 6, 8 10 is watched by 9, 11 Day 7: 5 is watched by 3, 4 8 is watched by 6, 7 11 is watched by 9, 10 This is the best I can come up with. In the later days, groups would be able to watch groups they've already seen of course. Eg, by day 7, groups 1 and 2 have seen every performance and been seen by every performance so they may as well go home. This answer doesn't feel quite as efficient as it could be so I figure someone might be able to find an answer for 6 days but that would be as low as it would go. Super Close! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 22, 2008 Report Share Posted December 22, 2008 (edited) Six seems to be the minimum Group1 performs on days 1 and 2 Group2 performs on days 1 and 3 Group3 performs on days 1 and 4 Group4 performs on days 1 and 5 Group5 performs on days 2 and 3 Group6 performs on days 2 and 4 Group7 performs on days 2 and 5 Group8 performs on days 3 and 4 Group9 performs on days 3 and 5 Group10 performs on days 4 and 5 Group11 performs on day 6 All groups are able to preform for all the others in six days. It was six. Nice job! Edited December 22, 2008 by Riddle Master Zack Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 22, 2008 Report Share Posted December 22, 2008 That solutions fits these answers: None of the the performances of a given day are simultaneous. No group performs more than once a day. No group performs and watches on the same day. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 22, 2008 Report Share Posted December 22, 2008 That solutions fits these answers: None of the the performances of a given day are simultaneous. No group performs more than once a day. No group performs and watches on the same day. Yes that's right sorry i didn't clarify. Quote Link to comment Share on other sites More sharing options...
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Eleven theater groups took part in a festival. Every day some of the groups put on their plays while the others watched.
When the festival was over it was possible to affirm that each one of the groups had been able to attend, at least once, the
performances of each of the other groups.
What is the minimum number of days that the festival lasted?
Good luck and have fun!-zak
Edited by Riddle Master ZackLink to comment
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