2016 AMC 8 Problems/Problem 18

Problem

In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?

$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$

Solution

Solution 1

From any $n-$ th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal. Starting with the first race: $\frac{216}{6}=36$ $\frac{36}{6}=6$ $\frac{6}{6}=1$ Adding all of the numbers in the second column yields $\boxed{\textbf{(C)}\ 43}$

Solution 2

Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. You can construct the equation: $216$ - $5x$ = $1$ . Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $\boxed{\textbf{(C)}\ 43}$

Solution 3 (Cheap Solution using Answer Choices)

Since $216$ is a power of $6,$ the answer will be in the form of $6x+1.$ We can see that this is odd and the only option is $\boxed{\textbf{(C)}\ 43}$ ~sanaops9

Video Solution

https://youtu.be/XIxOULink2I

~savannahsolver

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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