Difference between revisions of "2002 AMC 12B Problems/Problem 22"
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+ | === Solution 3 === | ||
+ | By the change of base formula, <math>\log_n 2002</math> and <math>\log_{2002} n</math> are reciprocals, so | ||
+ | <cmath>a_n = \frac{1}{\log_n 2002} = \log_{2002} n</cmath> | ||
+ | for all <math>n</math>. | ||
+ | |||
+ | Then,\begin{align*} | ||
+ | b - c &= (a_2 + a_3 + a_4 + a_5) - (a_{10} + a_{11} + a_{12} + a_{13} + a_{14}) \\ | ||
+ | &= (\log_{2002} 2 + \log_{2002} 3 + \log_{2002} 4 + \log_{2002} 5) \\ | ||
+ | &\quad - (\log_{2002} 10 + \log_{2002} 11 + \log_{2002} 12 + \log_{2002} 13 + \log_{2002} 14) \\ | ||
+ | &= \log_{2002} \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14} \\ | ||
+ | &= \log_{2002} \frac{1}{2002} \\ | ||
+ | &= \boxed{-1}. | ||
+ | \end{align*} | ||
+ | The answer is (B). | ||
== See also == | == See also == |
Revision as of 21:30, 23 November 2020
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
Solution 3
By the change of base formula, and are reciprocals, so for all .
Then,\begin{align*} b - c &= (a_2 + a_3 + a_4 + a_5) - (a_{10} + a_{11} + a_{12} + a_{13} + a_{14}) \\ &= (\log_{2002} 2 + \log_{2002} 3 + \log_{2002} 4 + \log_{2002} 5) \\ &\quad - (\log_{2002} 10 + \log_{2002} 11 + \log_{2002} 12 + \log_{2002} 13 + \log_{2002} 14) \\ &= \log_{2002} \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14} \\ &= \log_{2002} \frac{1}{2002} \\ &= \boxed{-1}. \end{align*} The answer is (B).
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 |
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