Difference between revisions of "2013 AMC 12A Problems/Problem 4"

 
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<math>\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}</math>, which is <math>C</math>
 
<math>\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}</math>, which is <math>C</math>
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==Solution 2==
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Suppose that <cmath>x=2^{2012}.</cmath> Then the given expression is equal to <cmath>\frac{4x+x}{4x-x}=\frac{5x}{3x}=\boxed{\frac{5}{3}\textbf{(C)}}.</cmath>
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~sugar_rush
  
 
==Video Solution==
 
==Video Solution==
https://www.youtu.be/2vf843cvVzo?t=545 (solution for #4 starts at 9:05)
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https://www.youtube.com/watch?v=2vf843cvVzo?t=545 (solution for #4 starts at 9:05)
  
 
~sugar_rush
 
~sugar_rush

Latest revision as of 22:13, 23 November 2020

Problem

What is the value of \[\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?\]

$\textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024}$

Solution

$\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}$

We can factor a ${2^{2012}}$ out of the numerator and denominator to obtain

$\frac{2^{2012}*(2^2+1)}{2^{2012}*(2^2-1)}$

The ${2^{2012}}$ cancels, so we get

$\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}$, which is $C$

Solution 2

Suppose that \[x=2^{2012}.\] Then the given expression is equal to \[\frac{4x+x}{4x-x}=\frac{5x}{3x}=\boxed{\frac{5}{3}\textbf{(C)}}.\]

~sugar_rush

Video Solution

https://www.youtube.com/watch?v=2vf843cvVzo?t=545 (solution for #4 starts at 9:05)

~sugar_rush

See also

2013 AMC 10A (ProblemsAnswer KeyResources)
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
2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 12 Problems and Solutions

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