Difference between revisions of "2002 AMC 12B Problems/Problem 22"
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&= \left(\frac{1}{\log 2002}\right)\left(\log \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}\right)\\ | &= \left(\frac{1}{\log 2002}\right)\left(\log \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}\right)\\ | ||
&= \left(\frac{1}{\log 2002}\right) \log 2002^{-1} = -\left(\frac{\log 2002}{\log 2002}\right) = -1 \Rightarrow \mathrm{(B)}\end{align*}</cmath> | &= \left(\frac{1}{\log 2002}\right) \log 2002^{-1} = -\left(\frac{\log 2002}{\log 2002}\right) = -1 \Rightarrow \mathrm{(B)}\end{align*}</cmath> | ||
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+ | == Solution 2 == | ||
+ | |||
+ | Note that <math>\frac{1}{\log_a b}=\log_b a</math>. Thus <math>a_n=\log_{2002} n</math>. Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following: | ||
+ | |||
+ | <cmath>\log_{2002}\left(\frac{2*3*4*5}{10*11*12*13*14}=\frac{1}{11*13*14}=\frac{1}{2002}\right)=\boxed{\textbf{(B)}-1}</cmath> | ||
+ | |||
+ | ~yofro | ||
== See also == | == See also == |
Revision as of 17:47, 26 October 2020
Contents
Problem
For all integers greater than , define . Let and . Then equals
Solution
By the change of base formula, . Thus
Solution 2
Note that . Thus . Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following:
~yofro
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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.