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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 9"
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Let <math>a_{n}</math> be a geometric sequence for <math>n\in\mathbb{Z}</math> with <math>a_{0}=1024</math> and <math>a_{10}=1</math>. Let <math>S</math> denote the infinite sum: <math>a_{10}+a_{11}+a_{12}+...</math>. If the sum of all distinct values of <math>S</math> is <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, then compute the sum of the positive prime factors of <math>n</math>.
 
Let <math>a_{n}</math> be a geometric sequence for <math>n\in\mathbb{Z}</math> with <math>a_{0}=1024</math> and <math>a_{10}=1</math>. Let <math>S</math> denote the infinite sum: <math>a_{10}+a_{11}+a_{12}+...</math>. If the sum of all distinct values of <math>S</math> is <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, then compute the sum of the positive prime factors of <math>n</math>.
  
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==Solution==
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==Solutions==
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===Solution 1===
 
Let the [[ratio]] of consecutive terms of the sequence be <math>r \in \mathbb{C}</math>.  Then we have by the given that <math>1 = a_{10} = r^{10} a_0 = 1024r^{10}</math> so <math>r^{10} = 2^{-10}</math> and <math>r = \frac \omega 2</math>, where <math>\omega</math> can be any of the tenth [[roots of unity]].
 
Let the [[ratio]] of consecutive terms of the sequence be <math>r \in \mathbb{C}</math>.  Then we have by the given that <math>1 = a_{10} = r^{10} a_0 = 1024r^{10}</math> so <math>r^{10} = 2^{-10}</math> and <math>r = \frac \omega 2</math>, where <math>\omega</math> can be any of the tenth [[roots of unity]].
  
 
Then the sum <math>S = a_{10} + a_{11} + \ldots = 1 + r + r^2 +\ldots = \frac{1}{1-r}</math> has value <math>\frac 1{1 - \omega / 2}</math>.  Different choices of <math>\omega</math> clearly lead to different values for <math>S</math>, so we don't need to worry about the distinctness condition in the problem.  Then the value we want is <math>\sum_{\omega^{10} = 1} \sum_{i = 10}^\infty 1024 \left(\frac\omega2\right)^i  = 1024 \sum_{i = 10}^\infty 2^{-i} \sum_{\omega^{10}=1} \omega^i</math>.  Now, recall that if <math>z_1, z_2, \ldots, z_n</math> are the <math>n</math> <math>n</math>th [[root of unity | roots of unity]] then for any [[integer]] <math>m</math>, <math>z_1^m + \ldots + z_n^m</math> is 0 unless <math>n | m</math> in which case it is 1.  Thus this simplifies to ...
 
Then the sum <math>S = a_{10} + a_{11} + \ldots = 1 + r + r^2 +\ldots = \frac{1}{1-r}</math> has value <math>\frac 1{1 - \omega / 2}</math>.  Different choices of <math>\omega</math> clearly lead to different values for <math>S</math>, so we don't need to worry about the distinctness condition in the problem.  Then the value we want is <math>\sum_{\omega^{10} = 1} \sum_{i = 10}^\infty 1024 \left(\frac\omega2\right)^i  = 1024 \sum_{i = 10}^\infty 2^{-i} \sum_{\omega^{10}=1} \omega^i</math>.  Now, recall that if <math>z_1, z_2, \ldots, z_n</math> are the <math>n</math> <math>n</math>th [[root of unity | roots of unity]] then for any [[integer]] <math>m</math>, <math>z_1^m + \ldots + z_n^m</math> is 0 unless <math>n | m</math> in which case it is 1.  Thus this simplifies to ...
  
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===Solution 2===
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Another solution:
 
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<math>\sum\frac1{1-z/2}</math> where <math>z^{10}=1</math>.
 
<math>\sum\frac1{1-z/2}</math> where <math>z^{10}=1</math>.
 
   
 
   
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==See Also==
 
==See Also==
  
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*[[Mock AIME 1 2006-2007/Problem 8 | Previous Problem]]
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*[[Mock AIME 1 2006-2007 Problems/Problem 8 | Previous Problem]]
  
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*[[Mock AIME 1 2006-2007/Problem 10 | Next Problem]]
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*[[Mock AIME 1 2006-2007 Problems/Problem 10 | Next Problem]]
  
 
*[[Mock AIME 1 2006-2007]]
 
*[[Mock AIME 1 2006-2007]]
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 14:51, 3 April 2012

Problem

Revised statement

Let $a_{n}$ be a geometric sequence of complex numbers with $a_{0}=1024$ and $a_{10}=1$, and let $S$ denote the infinite sum $S = a_{10}+a_{11}+a_{12}+...$. If the sum of all possible distinct values of $S$ is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, compute the sum of the positive prime factors of $n$.

Original statement

Let $a_{n}$ be a geometric sequence for $n\in\mathbb{Z}$ with $a_{0}=1024$ and $a_{10}=1$. Let $S$ denote the infinite sum: $a_{10}+a_{11}+a_{12}+...$. If the sum of all distinct values of $S$ is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, then compute the sum of the positive prime factors of $n$.

Solutions

Solution 1

Let the ratio of consecutive terms of the sequence be $r \in \mathbb{C}$. Then we have by the given that $1 = a_{10} = r^{10} a_0 = 1024r^{10}$ so $r^{10} = 2^{-10}$ and $r = \frac \omega 2$, where $\omega$ can be any of the tenth roots of unity.

Then the sum $S = a_{10} + a_{11} + \ldots = 1 + r + r^2 +\ldots = \frac{1}{1-r}$ has value $\frac 1{1 - \omega / 2}$. Different choices of $\omega$ clearly lead to different values for $S$, so we don't need to worry about the distinctness condition in the problem. Then the value we want is $\sum_{\omega^{10} = 1} \sum_{i = 10}^\infty 1024 \left(\frac\omega2\right)^i = 1024 \sum_{i = 10}^\infty 2^{-i} \sum_{\omega^{10}=1} \omega^i$. Now, recall that if $z_1, z_2, \ldots, z_n$ are the $n$ $n$th roots of unity then for any integer $m$, $z_1^m + \ldots + z_n^m$ is 0 unless $n | m$ in which case it is 1. Thus this simplifies to ...

Solution 2

$\sum\frac1{1-z/2}$ where $z^{10}=1$.

Let $t=\frac1{1-z/2}\implies z=2(1-1/t)$,

and $2^{10}(1-1/t)^{10}-1=0\implies2^{10}(t-1)^{10}-t^{10}=0$

We seek $\sum t$, or the negative of the coefficient of $t^9$ divided by the coefficient of $t^10$, which is $2^{10}\cdot10/(2^{10}-1)=2^{11}\cdot5/(2^{10}-1)$ and $2^{10}-1=33*31=3*11*31$.

Therefore the answer is 45.


See Also

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