Difference between revisions of "2016 AMC 10B Problems/Problem 2"
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| + | ==Problem== | ||
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| + | If <math>n\heartsuit m=n^3m^2</math>, what is <math>\frac{2\heartsuit 4}{4\heartsuit 2}</math>? | ||
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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> | ||
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==Solution 1== | ==Solution 1== | ||
<math>\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12</math> which is <math>\textbf{(B)}</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>. | ||
Revision as of 21:08, 28 January 2020
Contents
Problem
If 
, what is 
?
Solution 1
which is 
.
Solution 2
We can replace 
and 
with 
and 
respectively. Then substituting with 
and 
we can get 
and substitute to get 
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. 
