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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
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==Solution 3 (Elimination)==
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Choice <math>A</math> cannot be true, because <math>2^{2014}+2^{2012}</math> is clearly larger than <math>2^{2014}-2^{2012}</math>. We can apply our same logic to choice <math>B</math> and eliminate it as well. <math>2013</math> and <math>2^{4024}</math> are all whole numbers, but <math>2^{2014}+2^{2012}</math> is not a multiple of <math>2^{2014}-2^{2012}</math>, so we can eliminate choices <math>D</math> and <math>E</math> too. This leaves us with choice <math>\boxed{C}</math> as our final answer.
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~dbnl
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==Video Solution==
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https://www.youtube.com/watch?v=2vf843cvVzo?t=545 (solution for #4 starts at 9:05)
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 +
~sugar_rush
  
 
== See also ==
 
== See also ==
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{{AMC10 box|year=2013|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2013|ab=A|num-b=3|num-a=5}}
 
{{AMC12 box|year=2013|ab=A|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 12:49, 13 August 2023

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


Solution 3 (Elimination)

Choice $A$ cannot be true, because $2^{2014}+2^{2012}$ is clearly larger than $2^{2014}-2^{2012}$. We can apply our same logic to choice $B$ and eliminate it as well. $2013$ and $2^{4024}$ are all whole numbers, but $2^{2014}+2^{2012}$ is not a multiple of $2^{2014}-2^{2012}$, so we can eliminate choices $D$ and $E$ too. This leaves us with choice $\boxed{C}$ as our final answer.

~dbnl

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 (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
2013 AMC 12A (Problems • Answer Key • Resources)
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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