Difference between revisions of "2016 AMC 8 Problems/Problem 18"
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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 | ~sanaops9 | ||
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+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/OPC5GeUvwf8 | ||
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+ | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== |
Revision as of 16:57, 6 April 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: 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 (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
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.