Difference between revisions of "2003 AMC 10B Problems/Problem 15"
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===Solution 2=== | ===Solution 2=== | ||
| − | 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. Therefore, the answer is <math> | + | 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. Therefore, the answer is <math>\rightarrow\textbf{(E) }\text{divisible by 11.}</math> |
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Latest revision as of 22:23, 7 September 2023
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
Solution
Solution 1
Notice that 
players need to be eliminated for there to be declared a winner. Notice also that every match eliminates exactly one person. Therefore, matches are needed to eliminate people and therefore declare a winner. The rest of the information is irrelevant. Therefore, the total number of matches is 
.
Solution 2
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. Therefore, the answer is 
Video Solution by WhyMath
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
