Difference between revisions of "2014 AMC 10B Problems/Problem 2"
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Hence, the fraction equals to <math>\boxed{{64 (\textbf{E})}}</math>. | Hence, the fraction equals to <math>\boxed{{64 (\textbf{E})}}</math>. | ||
| + | ==Solution 2== | ||
| + | We could also do quick math\, solving the exponents to get: | ||
| + | <cmath>\frac{8+8}{</cmath>\frac{1}{8}+<cmath>\frac{1}{8}}\\=\frac{2^6+2^6}{1+1}\\={2^6}. </cmath> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:46, 24 July 2018
Contents
Problem
What is 
?
Solution
We can synchronously multiply 
to the polynomials both above and below the fraction bar. Thus,
![\[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\]](http://latex.artofproblemsolving.com/5/f/6/5f61712212b2338ec47df60099f80f0e4a58c157.png)
Hence, the fraction equals to 
.
Solution 2
We could also do quick math\, solving the exponents to get:
\[\frac{8+8}{\] (Error compiling LaTeX. Unknown error_msg)\frac{1}{8}+
\[\frac{1}{8}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\] (Error compiling LaTeX. Unknown error_msg)
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
