Difference between revisions of "2002 AMC 12B Problems/Problem 22"
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\qquad\mathrm{(D)}\ \frac{1}{1001} | \qquad\mathrm{(D)}\ \frac{1}{1001} | ||
\qquad\mathrm{(E)}\ \frac 12</math> | \qquad\mathrm{(E)}\ \frac 12</math> | ||
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== Solution == | == Solution == | ||
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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)\\ | ||
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&= \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 === | + | == Solution 2 == |
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+ | 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: | ||
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+ | <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> | ||
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+ | ~yofro | ||
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+ | == Solution 3 == | ||
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+ | Note that <math>a_2 = \frac{1}{\log_2 2002}</math>. <math>1</math> is also equal to <math>\log_2 2</math>. So <math>a_2 = \frac{\log_2 2}{\log_2 2002}</math>. By the change of bases formula, <math>a_2 = \log_{2002} 2</math>. Following the same reasoning, <math>a_3 = \log_{2002} 3</math>, <math>a_4 = \log_{2002} 4</math> and so on. | ||
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+ | <cmath>b = \log_{2002} 2 + \log_{2002} 3 + .....+ \log_{2002} 5 = \log_{2002} 5! = \log_{2002} 120</cmath> | ||
+ | Now solving for <math>c</math>, we see that it equals <math>\log_{2002} (10\cdot 11 \cdot 12 \cdot 13 \cdot 14)</math> | ||
+ | <cmath>b-c = \log_{2002} 120 - \log_{2002} 240240 \rightarrow \log_{2002} \frac{1}{2002} = \boxed{-1}</cmath> | ||
− | + | ~YBSuburbanTea | |
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== See also == | == See also == |
Latest revision as of 18:26, 13 January 2022
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
Note that .
is also equal to
. So
. By the change of bases formula,
. Following the same reasoning,
,
and so on.
Now solving for
, we see that it equals
~YBSuburbanTea
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.