Difference between revisions of "2016 AMC 10B Problems/Problem 2"
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<math>\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4</math> | <math>\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4</math> | ||
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==Solution== | ==Solution== | ||
− | <math>\ | + | <math>\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12</math> which is <math>\textbf{(B)}</math>. |
− | + | ==See Also== | |
+ | {{AMC10 box|year=2016|ab=B|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} |
Revision as of 10:49, 21 February 2016
Problem
If , what is ?
Solution
which is .
See Also
2016 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.