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>? | If <math>n\heartsuit m=n^3m^2</math>, what is <math>\frac{2\heartsuit 4}{4\heartsuit 2}</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> | <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== | ||
| + | <math>\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12</math> which is <math>\textbf{(B)}</math>. | ||
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| + | ==Solution 2== | ||
| + | We can replace <math>2</math> and <math>4</math> with <math>a</math> and <math>b</math> respectively. Then substituting with <math>n</math> and <math>m</math> we can get <math>\dfrac{a^3b^2}{b^3a^2}=\dfrac{a}{b}</math> and substitute to get <math>\dfrac{2}{4}=\boxed{\dfrac{1}{2}}</math> which is <math>\boxed{\textbf{(B)}}</math> | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/boCvD0Hb6h0 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
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| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/W7IwD6sZWco | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2016|ab=B|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:48, 2 July 2023
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 
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
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. 
