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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-We can factor a 2^2012 out of the numerator and denominator to obtain
+== Problem ==
-((2^2012)((2^2)+1))/((2^2012)((2^2)-1))
+What is the value of <cmath>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?</cmath>
-Both the (2^2012) terms cancel, so we get
+<math> \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} </math>
-((2^2)+1)/((2^2)-1) = 5/3, which is C
+==Solution==
+<math>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}</math>
+We can factor a <math>{2^{2012}}</math> out of the numerator and denominator to obtain
+<math>\frac{2^{2012}*(2^2+1)}{2^{2012}*(2^2-1)}</math>
+The <math>{2^{2012}}</math> cancels, so we get
+<math>\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}</math>, which is <math>C</math>
+==Solution 2==
+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>
+~sugar_rush
+==Solution 3 (Elimination)==
+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.
+~dbnl
+==Video Solution==
+https://www.youtube.com/watch?v=2vf843cvVzo?t=545 (solution for #4 starts at 9:05)
+~sugar_rush
+== See also ==
+{{AMC10 box|year=2013|ab=A|num-b=7|num-a=9}}
+{{AMC12 box|year=2013|ab=A|num-b=3|num-a=5}}
+{{MAA Notice}}

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