Difference between revisions of "2019 AMC 12A Problems/Problem 23"
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==Solution== | ==Solution== | ||
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| + | Using the recursive definition, <math>a_4 = (4 \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)</math> or <math>a_4 = (4^{m})^{n}</math> where <math>m = \frac{1}{\log_{7}(3)}</math> and <math>n = \log_{7}(3^{\frac{1}{\log_{7}(2)}})</math>. Using logarithm rules, we can remove the exponent of the 3 so that <math>n = \frac{\log_{7}(3)}{\log_{7}(2)}</math>. Therefore, <math>a_4 = 4^{\frac{1}{\log_{7}(2)}}</math>, which is <math>4 \, \heartsuit \, 2</math>. | ||
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| + | We claim that <math>a_n = n \, \heartsuit \, 2</math> for all <math>n \geq 3</math>. We can prove this through induction. | ||
==See Also== | ==See Also== | ||
Revision as of 18:04, 9 February 2019
Problem
Define binary operations 
and 
by ![\[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\]](http://latex.artofproblemsolving.com/5/f/0/5f0ef0b94787e769e936ea5ec8657b108b356c9e.png)
for all real numbers 
and 
for which these expressions are defined. The sequence 
is defined recursively by 
and ![\[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\]](http://latex.artofproblemsolving.com/f/1/b/f1b2d6d1783ff5064ff088b0b0ff162c3de58efc.png)
for all integers 
. To the nearest integer, what is 
?
Solution
Using the recursive definition, 
or 
where 
and 
. Using logarithm rules, we can remove the exponent of the 3 so that 
. Therefore, 
, which is 
.
We claim that 
for all 
. We can prove this through induction.
See Also
| 2019 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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. 
