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Difference between revisions of "1999 AIME Problems/Problem 13"

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==Solution==
 
==Solution==
  
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There are <math>{40 \choose 2} = 780</math> total pairings of teams, and thus <math>2^{780}</math> possible outcomes. In order for no two teams to win the same number of games, they must each win a different number of games. Since the minimum and maximum possible number of games won are 0 and 39 respectively, and there are 40 teams in total, each team corresponds uniquely with some <math>k</math>, with <math>0 \leq k \leq 39</math>, where <math>k</math> represents the number of games the team won. With this in mind, we see that there are a total of <math>40!</math> outcomes in which no two teams win the same number of games.  
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There are <math>{40 \choose 2} = 780</math> total pairings of teams, and thus <math>2^{780}</math> possible outcomes. In order for no two teams to win the same number of games, they must each win a different number of games. Since the minimum and maximum possible number of games won are 0 and 39 respectively, and there are 40 teams in total, each team corresponds uniquely with some <math>k</math>, with <math>0 \leq k \leq 39</math>, where <math>k</math> represents the number of games the team won. With this in mind, we see that there are a total of <math>40!</math> outcomes in which no two teams win the same number of games. Further, note that these are all the valid combinations, as the team with 1 win must beat the team with 0 wins, the team with 2 wins must beat the teams with 1 and 0 wins, and so on; thus, this uniquely defines a combination.
  
  
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Letting <math>y = \frac{40!}{2^{38}}</math>, we have that <math>\frac{40!}{2^{780}} = \frac{2^{38} y}{2^{780}} = \frac{y}{2^{742}}</math>. Since <math>y</math> and <math>2^{742}</math> are relatively prime, our quotient is in the desired form. Our answer is thus <math>\log_2 2^{742} = \boxed{742}</math>.
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<math>780-38 = \boxed{742}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 00:38, 6 October 2015

Problem

Forty teams play a tournament in which every team plays every other team exactly once. No ties occur, and each team has a $50 \%$ chance of winning any game it plays. The probability that no two teams win the same number of games is $\frac mn,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $\log_2 n.$

Solution

There are ${40 \choose 2} = 780$ total pairings of teams, and thus $2^{780}$ possible outcomes. In order for no two teams to win the same number of games, they must each win a different number of games. Since the minimum and maximum possible number of games won are 0 and 39 respectively, and there are 40 teams in total, each team corresponds uniquely with some $k$, with $0 \leq k \leq 39$, where $k$ represents the number of games the team won. With this in mind, we see that there are a total of $40!$ outcomes in which no two teams win the same number of games. Further, note that these are all the valid combinations, as the team with 1 win must beat the team with 0 wins, the team with 2 wins must beat the teams with 1 and 0 wins, and so on; thus, this uniquely defines a combination.


The desired probability is thus $\frac{40!}{2^{780}}$. We wish to simplify this into the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime. The only necessary step is to factor out all the powers of 2 from $40!$; the remaining number is clearly relatively prime to all powers of 2.


The number of powers of 2 in $40!$ is $\left \lfloor \frac{40}{2} \right \rfloor + \left \lfloor \frac{40}{4} \right \rfloor + \left \lfloor \frac{40}{8} \right \rfloor + \left \lfloor \frac{40}{16} \right \rfloor + \left \lfloor \frac{40}{32} \right \rfloor = 20 + 10 + 5 + 2 + 1 = 38.$


$780-38 = \boxed{742}$.

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AIME Problems and Solutions

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