Difference between revisions of "2019 AMC 12A Problems/Problem 23"
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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. | We claim that <math>a_n = n \, \heartsuit \, 2</math> for all <math>n \geq 3</math>. We can prove this through induction. | ||
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+ | <math>a_n = a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, ((n-1) \, \heartsuit \, 2)</math> | ||
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+ | This can be simplified as <math>a_n = ((n^{\log_{n-1}(7)}) \, \diamondsuit \, ((n-1)^{\log_{2}(7)}))</math>. | ||
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+ | Applying the diamond operation, we can simplify <math>a_n = n^h</math> where <math>h = \log_{n-1}(7) \cdot \log_{7}(n-1)^{\log_{2}(7)}</math>. By using logarithm rules to remove the exponent of <math>\log_{7}(n-1)</math> and after cancelling, <math>h = \frac{1}{\log_{7}(2)}</math>. | ||
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+ | Therefore, <math>a_n = n^{\frac{1}{\log_{7}(2)}} = n \, \heartsuit \, 2</math> for all <math>n \geq 3</math>, completing the induction. | ||
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+ | We have <math>a_2019 = 2019^{\log_{2}(7)}</math>. Taking log base 2019 of both sides gives us <math>{\log_{2019}(a_{2019})} = {\log_{2}(7)}</math>. Then, by changing to base 7 and after cancellation, we arrive at <math>{\log_{7}(a_{2019})} = {\log_{2}(2019)}</math>. Because <math>2^{11} = 2048</math> and <math>2^{10} = 1024</math>, our answer is <math>\boxed{\textbf{(D)}11}</math> | ||
==See Also== | ==See Also== |
Revision as of 20:46, 9 February 2019
Contents
Problem
Define binary operations and by for all real numbers and for which these expressions are defined. The sequence is defined recursively by and for all integers . To the nearest integer, what is ?
Solution 1
By definition, the recursion becomes . By the change of base formula, this reduces to . Thus, we have . Thus, for each positive integer , the value of must be some constant value .
We now compute from . It is given that , so .
Now, we must have . Changing bases to , this becomes , so , where the last equality comes from the logarithmic chain rule. We conclude that , or choice .
Solution 2
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.
This can be simplified as .
Applying the diamond operation, we can simplify where . By using logarithm rules to remove the exponent of and after cancelling, .
Therefore, for all , completing the induction.
We have . Taking log base 2019 of both sides gives us . Then, by changing to base 7 and after cancellation, we arrive at . Because and , our answer is
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.