Difference between revisions of "2000 AIME II Problems/Problem 1"
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<cmath>\frac{2}{\log_4 2000^6} + \frac{3}{\log_5 2000^6}</cmath> <cmath>= \frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000}</cmath> <cmath>= \frac{1}{3} \log_{2000} 4 + \frac{1}{2} \log_{2000} 5</cmath> <cmath>= \log_{2000} (\sqrt[3]{4} \cdot \sqrt{5}) = x</cmath> <cmath>\implies 2^{4x} \cdot 5^{3x} = 2^{\frac{2}{3}} \cdot 5^{\frac{1}{2}}</cmath> <cmath>\implies 4x + (3\log_2 5)x = \frac{2}{3}+\frac{1}{2} \log_2 5</cmath> <cmath>\implies x = \frac{\frac{2}{3} + \frac{1}{2} \log_2 5}{4 + 3\log_2 5}</cmath> <cmath>\implies 6x = \frac{4 + 3\log_2 5}{4 + 3\log_2 5}</cmath> <cmath>\implies x = \frac{1}{6}</cmath> <cmath>\implies m + n = \boxed{007}</cmath> | <cmath>\frac{2}{\log_4 2000^6} + \frac{3}{\log_5 2000^6}</cmath> <cmath>= \frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000}</cmath> <cmath>= \frac{1}{3} \log_{2000} 4 + \frac{1}{2} \log_{2000} 5</cmath> <cmath>= \log_{2000} (\sqrt[3]{4} \cdot \sqrt{5}) = x</cmath> <cmath>\implies 2^{4x} \cdot 5^{3x} = 2^{\frac{2}{3}} \cdot 5^{\frac{1}{2}}</cmath> <cmath>\implies 4x + (3\log_2 5)x = \frac{2}{3}+\frac{1}{2} \log_2 5</cmath> <cmath>\implies x = \frac{\frac{2}{3} + \frac{1}{2} \log_2 5}{4 + 3\log_2 5}</cmath> <cmath>\implies 6x = \frac{4 + 3\log_2 5}{4 + 3\log_2 5}</cmath> <cmath>\implies x = \frac{1}{6}</cmath> <cmath>\implies m + n = \boxed{007}</cmath> | ||
Revision as of 14:56, 11 February 2024
Problem
The number
can be written as where and are relatively prime positive integers. Find .
Solution
Solution 1
Therefore,
Solution 2
Alternatively, we could've noted that, because
Therefore our answer is .
Solution 3
We know that and , and by base of change formula, . Lastly, notice for all bases.
~
Solution 4
~ cxsmi
2000 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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