2003 AMC 10B Problems/Problem 15

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

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

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

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

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

$\qquad\textbf{(E) } \text{divisible by 11}$

Solution

Solution 1

Notice that $99$ players need to be eliminated for there to be declared a winner. Notice also that every match eliminates exactly one person. Therefore, $99$ matches are needed to eliminate $99$ people and therefore declare a winner. The rest of the information is irrelevant. Therefore, the total number of matches is $\text{divisible by 11}, \boxed{\text{E}}$.

Solution 2

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. Therefore, the answer is $\rightarrow\textbf{(E) }\text{divisible by 11.}$

Video Solution by WhyMath

https://youtu.be/v-x_BO5aQDw

~savannahsolver

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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