Difference between revisions of "2016 AMC 8 Problems/Problem 18"
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===Solution 2=== | ===Solution 2=== | ||
− | Every race eliminates <math>5</math> players. The winner is decided when there is only <math>1</math> runner left. Thus, <math>215</math> players have to be eliminated. Therefore, we need <math>\frac{215}{5}</math> games to decide the winner, or <math>\boxed{\textbf{(C)}\ 43}</math> | + | Every race eliminates <math>5</math> players. The winner is decided when there is only <math>1</math> runner left. You can construct the equation: <math>216</math> - <math>5x</math> = <math>1</math>. Thus, <math>215</math> players have to be eliminated. Therefore, we need <math>\frac{215}{5}</math> games to decide the winner, or <math>\boxed{\textbf{(C)}\ 43}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2016|num-b=17|num-a=19}} | {{AMC8 box|year=2016|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:39, 26 December 2020
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
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 |
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