Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 9"

Revision as of 00:03, 5 November 2009

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 $n$ .

Solution

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 ...

Another solution:

$\sum\frac1{1-z/2}$ 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 $t^9$ , which is $2^10\cdot10=2^11\cdot5$ .

Therefore the answer is 2+5=7

@@ Line 10: / Line 10: @@
 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 ...
+Another solution:
+<math>\sum\frac1{1-z/2}</math> where <math>z^10=1</math>.
+Let <math>t=\frac1{1-z/2}\implies z=2(1-1/t)</math>,
+and <math>2^10(1-1/t)^10-1=0\implies2^10(t-1)^10-t^10=0</math>
+We seek <math>\sum t</math>, or the negative of the coefficient of <math>t^9</math>, which is <math>2^10\cdot10=2^11\cdot5</math>.
+Therefore the answer is 2+5=7

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