Difference between revisions of "2016 AMC 10B Problems/Problem 2"

(Solution)
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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>
 
 
  
 
==Solution==
 
==Solution==
<math>\textbf{(B)}\ \frac{1}{2}</math>
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<math>\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12</math> which is <math>\textbf{(B)}</math>.
  
Solution by ngeorge
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==See Also==
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{{AMC10 box|year=2016|ab=B|num-b=1|num-a=3}}
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{{MAA Notice}}

Revision as of 10:49, 21 February 2016

Problem

If $n\heartsuit m=n^3m^2$, what is $\frac{2\heartsuit 4}{4\heartsuit 2}$?

$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4$

Solution

$\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12$ which is $\textbf{(B)}$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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

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