Bump this,any solutions.

Bump, is the answer 4?

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Could the answer not technically be one if you cram all games into a day? How is the splitting of days decided?

I think answer is $ n-1 $ for n. I have got long solution for this problem when $ n=5 $ . Has Anybody short solutions?

There are 10 pairs of $ { A, B, C, D, E } $. Let $ {A,B}=1 {A,C}=2 {A,D}=3 {A,E}=4 {B,C}=5 {B,D}=6 {B,E}=7 {C,D}=8 {C,E}=9 {D,E}=10 $ It's clearly each number can play with only 3 numbers. ( For example: 1 can play with 8,9,10). Let's make a set for each number which this number can plays with numbers in his set. $ S_1={8,9,10}, S_2={6,10,7}, S_3={5,7,9} S_4={5,6,8} S_5={3,4,10} S_6={2,4,9} S_7={2,3,8} S_8={1,4,7} S_9={1,3,6} S_10={1,2,5} $. $ S_i $ is the set of i-th number. It's clearly total of games=15. It's clearly in one day at most 5 games. Because they're 10 teams and each team can play at most one game at per a day. By Implies this claims, we can easily find number of days $ >2 $ Example for 4: first day: (1,10), (2,7), (3,9), (4,6) second day: (1,9), (2,6), (3,7), (4,8), (5,10) third day: (1,8), (2,10), (3,5), (6,9) fourth day: (4,5), (7,8). $ (x,y) $ means x and y teams played. and what if number of days=3??? then once day has exactly 5 games: team 1 should play with their 3 opponents 8 9 and 10 at each one day. W.L.O.G First day 1 and 8 played. Second day 1 and 9 played. Third day 1 and 10 played. 1,2,3,4 cannot plays with each other. Then each day has 2 3 4 teams and they're also plays with each his 1 opponent at once day. So Now we know about 4 games at one day. 5-game??? We have got games: (6,9) (5,10) and (7,8). Game (6,9) cannot be participate at second day, Because 1 and 9 teams is playing at second day. So We have got 2 cases: CASE 1: (6,9) at first day, (5,10) at second day, (7,8) at third day. team 2 vs team 10. and this game should be in first day Similarly (2,7) game should be in second day and (2,6) game should be in third day. team 4 vs team 6. this game should be in second day. Because team 6 exist in first and third days. Similarly (4,8) game should be in second day but team 4 cannot play 2 games in one day. Contradiction CASE 2: (5,10) at first day, (7,8) at second day, (6,9) at third day. We can find contradiction similarly to first case. team 2 vs team 10 cannot be in the first day and cannot be in the third day. So (2,10) game should be in the second day. So by implies this statement we have (2,6) in the first day and (2,7) in the third day. (2,6) in the first day then We have (4,6) should be in the second day. (4,5) should be in the third day and (4,8) in the first day but team 8 playing with team 1 in the first day. It gives us Contradiction. Then Answer: $ 4 $.