Difference between revisions of "2017 AMC 12A Problems/Problem 25"
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<math>\textbf{(A) } \dfrac{5\cdot11}{3^{10}} \qquad \textbf{(B) } \dfrac{5^2\cdot11}{2\cdot3^{10}} \qquad \textbf{(C) } \dfrac{5\cdot11}{3^{9}} \qquad \textbf{(D) } \dfrac{5\cdot7\cdot11}{2\cdot3^{10}} \qquad \textbf{(E) } \dfrac{2^2\cdot5\cdot11}{3^{10}}</math> | <math>\textbf{(A) } \dfrac{5\cdot11}{3^{10}} \qquad \textbf{(B) } \dfrac{5^2\cdot11}{2\cdot3^{10}} \qquad \textbf{(C) } \dfrac{5\cdot11}{3^{9}} \qquad \textbf{(D) } \dfrac{5\cdot7\cdot11}{2\cdot3^{10}} \qquad \textbf{(E) } \dfrac{2^2\cdot5\cdot11}{3^{10}}</math> | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2017|ab=A|num-b=24|num-a=??}} | ||
| + | {{MAA Notice}} | ||
Revision as of 18:00, 8 February 2017
Problem
The vertices 
of a centrally symmetric hexagon in the complex plane are given by
![\[V=\left\{ \sqrt{2}i,-\sqrt{2}i, \frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.\]](http://latex.artofproblemsolving.com/2/4/5/245a3e5ef97921f9aacfa8b49f29b5ba1a176939.png)
For each 
, 
, an element 
is chosen from at random, independently of the other choices. Let 
be the product of the 
numbers selected. What is the probability that 
?
See Also
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Problem ?? |
| 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. 
