Difference between revisions of "2013 AMC 10A Problems/Problem 8"
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==Problem== | ==Problem== | ||
− | What is the value of < | + | 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== | ||
+ | |||
+ | Factoring out, we get: <math>\frac{2^{2012}(2^2 + 1)}{2^{2012}(2^2-1)}</math>. | ||
+ | |||
+ | Cancelling out the <math>2^{2012}</math> from the numerator and denominator, we see that it simplifies to <math>\boxed{\textbf{(C) }\frac{5}{3}}</math>. | ||
+ | ==Solution 2== | ||
+ | Let <math>x=2^{2012}</math> | ||
− | <math> \ | + | Then the given expression is equal to <math>\frac{4x+x}{4x-x}=\frac{5x}{3x}=\boxed{\textbf{(C) }\frac{5}{3}}</math> |
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/1ULw0x9XK4o | ||
+ | ~Education, the Study of Everything | ||
− | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=2vf843cvVzo?t=545 | ||
+ | ~sugar_rush | ||
− | + | https://youtu.be/C3prgokOdHc | |
− | + | ~savannahsolver | |
==See Also== | ==See Also== |
Latest revision as of 13:08, 1 July 2023
Contents
Problem
What is the value of
Solution
Factoring out, we get: .
Cancelling out the from the numerator and denominator, we see that it simplifies to
.
Solution 2
Let
Then the given expression is equal to
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=2vf843cvVzo?t=545
~sugar_rush
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.