Difference between revisions of "2002 AMC 12B Problems/Problem 22"

(Solution 2)
(Solution 3)
 
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== Solution 3 ==
 
== Solution 3 ==
  
Note that <math>a_2 = \frac{1}{\log_2 2022}</math>. <math>1</math> is also equal to <math>\log_2 2</math>. So <math>a_2 = \frac{\log_2 2}{\log_2 2022}</math>. By the change of bases formula, <math>a_2 = \log_{2022} 2</math>. Following the same reasoning, <math>a_3 = \log_{2022} 3</math>, <math>a_4 = \log_{2022} 4</math> and so on.
+
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.
  
<cmath>b = \log_{2022} 2 + \log_{2022} 3 + .....+ \log_{2022} 5 = \log_{2022} 5! = \log_{2022} 120</cmath>
+
<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_{2022} (10\cdot 11 \cdot 12 \cdot 13 \cdot 14)</math>
+
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_{2022} 120 - \log_{2022} 240240 \rightarrow \log_{2022} \frac{1}{2022} = \boxed{-1}</cmath>
+
<cmath>b-c = \log_{2002} 120 - \log_{2002} 240240 \rightarrow \log_{2002} \frac{1}{2002} = \boxed{-1}</cmath>
  
 
~YBSuburbanTea
 
~YBSuburbanTea

Latest revision as of 18:26, 13 January 2022

Problem

For all integers $n$ greater than $1$, define $a_n = \frac{1}{\log_n 2002}$. Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$. Then $b- c$ equals

$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1  \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \frac 12$

Solution

By the change of base formula, $a_n = \frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n$. Thus \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) \log 2002^{-1} = -\left(\frac{\log 2002}{\log 2002}\right) = -1 \Rightarrow \mathrm{(B)}\end{align*}

Solution 2

Note that $\frac{1}{\log_a b}=\log_b a$. Thus $a_n=\log_{2002} n$. 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:

\[\log_{2002}\left(\frac{2*3*4*5}{10*11*12*13*14}=\frac{1}{11*13*14}=\frac{1}{2002}\right)=\boxed{\textbf{(B)}-1}\]

~yofro

Solution 3

Note that $a_2 = \frac{1}{\log_2 2002}$. $1$ is also equal to $\log_2 2$. So $a_2 = \frac{\log_2 2}{\log_2 2002}$. By the change of bases formula, $a_2 = \log_{2002} 2$. Following the same reasoning, $a_3 = \log_{2002} 3$, $a_4 = \log_{2002} 4$ and so on.

\[b = \log_{2002} 2 + \log_{2002} 3 + .....+ \log_{2002} 5 = \log_{2002} 5! = \log_{2002} 120\] Now solving for $c$, we see that it equals $\log_{2002} (10\cdot 11 \cdot 12 \cdot 13 \cdot 14)$ \[b-c = \log_{2002} 120 - \log_{2002} 240240 \rightarrow \log_{2002} \frac{1}{2002} = \boxed{-1}\]

~YBSuburbanTea

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
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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