Difference between revisions of "2016 AMC 8 Problems/Problem 18"
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===Solution 3 (Cheap Solution using Answer Choices)=== | ===Solution 3 (Cheap Solution using Answer Choices)=== | ||
Since <math>216</math> is a power of <math>6,</math> the answer will be in the form of <math>6x+1.</math> We can see that this is odd and the only option is <math>\boxed{\textbf{(C)}\ 43}</math> | Since <math>216</math> is a power of <math>6,</math> the answer will be in the form of <math>6x+1.</math> We can see that this is odd and the only option is <math>\boxed{\textbf{(C)}\ 43}</math> | ||
| + | ~sanaops9 | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 14:13, 13 January 2023
Contents
Problem
In an All-Area track meet, 
sprinters enter a 
meter dash competition. The track has 
lanes, so only 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?
Solution
Solution 1
From any 
th race, only 
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:
![\[\frac{216}{6}=36\]](http://latex.artofproblemsolving.com/7/3/7/7374e5d18f64255ab216a8e82a352a114b66f4fd.png)
![\[\frac{36}{6}=6\]](http://latex.artofproblemsolving.com/a/9/1/a916513079199978782d4485856c6baa38221974.png)
![\[\frac{6}{6}=1\]](http://latex.artofproblemsolving.com/8/8/4/884f0b91255d6b75d42375947ec42a5e307d0bf9.png)
Adding all of the numbers in the second column yields 
Solution 2
Every race eliminates 
players. The winner is decided when there is only 
runner left. You can construct the equation: - 
= . Thus, 
players have to be eliminated. Therefore, we need 
games to decide the winner, or
Solution 3 (Cheap Solution using Answer Choices)
Since is a power of 
the answer will be in the form of 
We can see that this is odd and the only option is
~sanaops9
Video Solution
~savannahsolver
See Also
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
