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Difference between revisions of "2006 AIME I Problems/Problem 9"

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== Problem ==
 
== Problem ==
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The sequence <math> a_1, a_2, \ldots </math> is geometric with <math> a_1=a </math> and common ratio <math> r, </math> where <math> a </math> and <math> r </math> are positive integers. Given that <math> \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, </math> find the number of possible ordered pairs <math> (a,r). </math>
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The [[sequence]] <math> a_1, a_2, \ldots </math> is [[geometric sequence|geometric]] with <math> a_1=a </math> and common [[ratio]] <math> r, </math> where <math> a </math> and <math> r </math> are positive integers. Given that <math> \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, </math> find the number of possible ordered pairs <math> (a,r). </math>
  
 
== Solution ==
 
== Solution ==
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<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) = </math>
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<cmath>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) \\
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= \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8  (a^{12}r^{66}) </cmath>
  
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<math> = \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8  (a^{12}r^{66}) </math>
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So our question is equivalent to solving <math>\log_8 (a^{12}r^{66})=2006</math> for <math>a, r</math> [[positive integer]]s. <math>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so <math>a^{2}r^{11}=2^{1003}</math>.
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So our question is equivalent to solving <math>\log_8 (a^{12}r^{66})=2006</math> for <math>a, r</math> [[positive integer]]s.
 
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<math>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so
 
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<math>a^{2}r^{11}=2^{1003}</math>
 
  
 
The product of <math>a^2</math> and <math>r^{11}</math> is a power of 2.  Since both numbers have to be integers, this means that <math>a</math> and <math>r</math> are themselves powers of 2.  Now, let <math>a=2^x</math> and <math>r=2^y</math>:
 
The product of <math>a^2</math> and <math>r^{11}</math> is a power of 2.  Since both numbers have to be integers, this means that <math>a</math> and <math>r</math> are themselves powers of 2.  Now, let <math>a=2^x</math> and <math>r=2^y</math>:
  
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<math>(2^x)^2\cdot(2^y)^{11}=2^{1003}</math>
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<cmath>\begin{eqnarray*}(2^x)^2\cdot(2^y)^{11}&=&2^{1003}\\
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2^{2x}\cdot 2^{11y}&=&2^{1003}\\
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<math>2^{2x}\cdot 2^{11y}=2^{1003}</math>
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2x+11y&=&1003\\
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y&=&\frac{1003-2x}{11} \end{eqnarray*}</cmath>
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<math>2^{2x+11y}=2^{1003}</math>
 
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<math>2x+11y=1003</math>
 
  
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<math>11y=1003-2x</math>
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For <math>y</math> to be an integer, the [[numerator]] must be [[divisible]] by <math>11</math>.  This occurs when <math>x=1</math> because <math>1001=91*11</math>. Because only [[even integer]]s are being subtracted from <math>1003</math>, the numerator never equals an even [[multiple]] of <math>11</math>. Therefore, the numerator takes on the value of every [[odd integer | odd]] multiple of <math>11</math> from <math>11</math> to <math>1001</math>.  Since the odd multiples are separated by a distance of <math>22</math>, the number of ordered pairs that work is <math>1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46</math>.  (We must add 1 because both endpoints are being included.) So the answer is <math>\boxed{046}</math>.
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<math>y=\frac{1003-2x}{11}</math>
 
−
 
 
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For <math>y</math> to be an integer, the [[numerator]] must be [[divisible]] by 11.  This occurs when <math>x=1</math> because <math>1001=91*11</math>. Because only [[even integer]]s are being subtracted from 1003, the numerator never equals an even [[multiple]] of 11. Therefore, the numerator takes on the value of every [[odd integer | odd]] multiple of 11 from 11 to 1001.  Since the odd multiples are separated by a distance of 22, the number of ordered pairs that work is <math>1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46</math>.  (We must add 1 because both endpoints are being included.) So the answer is <math>46</math>.
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2006|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2006|n=I|num-b=8|num-a=10}}
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−
[[Category:Intermediate Geometry Problems]]
 
−
[[Category:Intermediate Trigonometry Problems]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
−
 
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 18:52, 28 November 2007

Problem

The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$

Solution

\[\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) \\ = \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8 (a^{12}r^{66})\]

So our question is equivalent to solving $\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers. $a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}$ so $a^{2}r^{11}=2^{1003}$.

The product of $a^2$ and $r^{11}$ is a power of 2. Since both numbers have to be integers, this means that $a$ and $r$ are themselves powers of 2. Now, let $a=2^x$ and $r=2^y$:

\begin{eqnarray*}(2^x)^2\cdot(2^y)^{11}&=&2^{1003}\\ 2^{2x}\cdot 2^{11y}&=&2^{1003}\\ 2x+11y&=&1003\\ y&=&\frac{1003-2x}{11} \end{eqnarray*}

For $y$ to be an integer, the numerator must be divisible by $11$. This occurs when $x=1$ because $1001=91*11$. Because only even integers are being subtracted from $1003$, the numerator never equals an even multiple of $11$. Therefore, the numerator takes on the value of every odd multiple of $11$ from $11$ to $1001$. Since the odd multiples are separated by a distance of $22$, the number of ordered pairs that work is $1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46$. (We must add 1 because both endpoints are being included.) So the answer is $\boxed{046}$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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