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. | |
==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 22:53, 14 December 2011
Problem
There are 
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 
players are given a bye, and the remaining 
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. Unknown error_msg)
Solution
28 people receive byes, so in the first round there are 
matches played. In the second round there are 
people, so there are 32 matches. In the subsequent rounds, there are 
matches played, for a total of 
matches. Divisible by 11.
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
