Difference between revisions of "2002 AMC 12B Problems/Problem 22"
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\qquad\mathrm{(E)}\ \frac 12</math> | \qquad\mathrm{(E)}\ \frac 12</math> | ||
== Solution == | == Solution == | ||
| + | === Solution 1 === | ||
By the [[Logarithms#Logarithmic Properties|change of base]] formula, <math>a_n = \frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n </math>. Thus | By the [[Logarithms#Logarithmic Properties|change of base]] formula, <math>a_n = \frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n </math>. Thus | ||
<cmath>\begin{align*}b- c &= \left(\frac{1}{\log 2002}\right)(\log 2 + \log 3 + \log 4 + \log 5 - \log 10 - \log 11 - \log 12 - \log 13 - \log 14)\\ | <cmath>\begin{align*}b- c &= \left(\frac{1}{\log 2002}\right)(\log 2 + \log 3 + \log 4 + \log 5 - \log 10 - \log 11 - \log 12 - \log 13 - \log 14)\\ | ||
&= \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> | ||
| + | |||
| + | === Solution 2 === | ||
| + | |||
| + | <math>a_n = \frac{1}{\log_n 2002} = \frac{\log_n n}{log_n 2002} = \log_2002 n</math> So | ||
| + | <cmath>\begin{align*} | ||
| + | b- c &= \left(log_2002 2 + log_2002 3 + log_2002 4 + log_2002 5 - log_2002 10 - log_2002 11 - log_2002 12 - log_2002 13 - log_2002 14))\\ | ||
| + | &= \left(log_2002 (\frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}))\\ | ||
| + | &= \left(log_2002 2002^-1) = -1 \Rightarrow \mathrm{(B)} | ||
| + | \end{align*}</cmath> | ||
== See also == | == See also == | ||
Revision as of 22:32, 26 November 2017
Problem
For all integers 
greater than 
, define 
. Let 
and 
. Then 
equals
Solution
Solution 1
By the change of base formula, 
. Thus
Solution 2
So
\begin{align*}
b- c &= \left(log_2002 2 + log_2002 3 + log_2002 4 + log_2002 5 - log_2002 10 - log_2002 11 - log_2002 12 - log_2002 13 - log_2002 14))\\
&= \left(log_2002 (\frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}))\\
&= \left(log_2002 2002^-1) = -1 \Rightarrow \mathrm{(B)}
\end{align*} (Error compiling LaTeX. Unknown error_msg)
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. 
