Difference between revisions of "2016 AMC 10A Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | If you started backwards you would get: <cmath>0\Rightarrow (+40)=40 , \Rightarrow (\frac{1}{2})=20 , \Rightarrow (+40)=60 , \Rightarrow (\frac{1}{2})=30 , \Rightarrow (+40)=70 , \Rightarrow (\frac{1}{2})=\boxed{\textbf{(C) }35}</cmath> | + | If you started backwards you would get: <cmath>0\Rightarrow (+40)=40 , \Rightarrow \left(\frac{1}{2}\right)=20 , \Rightarrow (+40)=60 , \Rightarrow \left(\frac{1}{2}\right)=30 , \Rightarrow (+40)=70 , \Rightarrow \left(\frac{1}{2}\right)=\boxed{\textbf{(C) }35}</cmath> |
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| + | == Solution 2 == | ||
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| + | If you have <math>x</math> as the amount of money Foolish Fox started with we have <math>2(2(2x-40)-40)-40=0.</math> Solving this we get <math>\boxed{\textbf{(C) }35}</math>. | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 13:59, 8 August 2021
Problem
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays 
coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?
Solution
If you started backwards you would get: ![\[0\Rightarrow (+40)=40 , \Rightarrow \left(\frac{1}{2}\right)=20 , \Rightarrow (+40)=60 , \Rightarrow \left(\frac{1}{2}\right)=30 , \Rightarrow (+40)=70 , \Rightarrow \left(\frac{1}{2}\right)=\boxed{\textbf{(C) }35}\]](http://latex.artofproblemsolving.com/3/5/b/35bfad9a40fc66594633eb4068a13dbe6125a1ab.png)
Solution 2
If you have 
as the amount of money Foolish Fox started with we have 
Solving this we get 
.
Video Solution
https://youtu.be/XXX4_oBHuGk?t=323
~IceMatrix
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
