Difference between revisions of "2003 AMC 10B Problems/Problem 15"

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==Solution==
 
==Solution==
 
   
 
   
28 people receive byes, so in the first round there are <math>36</math> matches played. In the second round there are <math>28 + 36 = 64</math> people, so there are 32 matches. In the subsequent rounds, there are <math>16, 8, 4, 2, 1</math> matches played, for a total of <math>36 + 32 + 16 + 8 + 4 + 2 + 1 = 99</math> matches. Divisible by 11.
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28 people receive byes, so in the first round there are <math>36</math> matches played. In the second round there are <math>28 + 36 = 64</math> people, so there are 32 matches. In the subsequent rounds, there are <math>16, 8, 4, 2, 1</math> matches played, for a total of <math>36 + 32 + 16 + 8 + 4 + 2 + 1 = 99</math> matches. Divisible by 11 <math>\Rightarrow</math> <math>\boxed{(E)} </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2003|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2003|ab=B|num-b=14|num-a=16}}
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Revision as of 13:54, 29 December 2011

Problem

There are $100$ players in a single tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest $28$ players are given a bye, and the remaining $72$ players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played is

$\textbf{(A) } \text{a prime number}

\qquad\textbf{(B) } \text{divisible by 2}

\qquad\textbf{(C) } \text{divisible by 5}

\qquad\textbf{(D) } \text{divisible by 7}

\qquad\textbf{(E) } \text{divisible by 11}$ (Error compiling LaTeX. ! Missing $ inserted.)

Solution

28 people receive byes, so in the first round there are $36$ matches played. In the second round there are $28 + 36 = 64$ people, so there are 32 matches. In the subsequent rounds, there are $16, 8, 4, 2, 1$ matches played, for a total of $36 + 32 + 16 + 8 + 4 + 2 + 1 = 99$ matches. Divisible by 11 $\Rightarrow$ $\boxed{(E)}$.

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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