Difference between revisions of "2013 AMC 12A Problems/Problem 4"
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+ | == Problem == | ||
+ | |||
+ | What is the value of <cmath>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?</cmath> | ||
+ | |||
+ | <math> \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | |||
<math>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}</math> | <math>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}</math> | ||
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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> | ||
+ | |||
+ | ==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}} |
Latest revision as of 13:49, 13 August 2023
Problem
What is the value of
Solution
We can factor a out of the numerator and denominator to obtain
The cancels, so we get
, which is
Solution 2
Suppose that Then the given expression is equal to
~sugar_rush
Solution 3 (Elimination)
Choice cannot be true, because
is clearly larger than
. We can apply our same logic to choice
and eliminate it as well.
and
are all whole numbers, but
is not a multiple of
, so we can eliminate choices
and
too. This leaves us with choice
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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