Mock AIME 1 2006-2007 Problems/Problem 9

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

$\sum\frac1{1-z/2}$ 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 $\boxed{45}$ .

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