# Difference between revisions of "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 $\boxed{45}$.

### Solution 2

To answer the original problem statement, let the common geometric be $r$ and $a_n$ denotes a term in the sequence. The term $a_{10}$ can be represented as $a\cdot r^{10}$ and this expression has to be set equal to one. By simplifying, we get $(2r)^{10}=1$ where the polynomial $r^{10}$ can have ten possible values or roots. Now, we take a look at the needed infinite geometric sum. The expression $a_{10}+a_{11}+a_{12}+...$ can be expressed as $1+r+r^2+r^3+...$ or even more better, $\frac{1}{1-r}$. However, the question is asking for the sum of the distinct roots of the polynomial, and as stated before, $r^{10}$ is a tenth-degree polynomial and it had ten different roots. Therefore, our desired sum is $\frac{1}{1-r_{1}}+\frac{1}{1-r_{2}}+\frac{1}{1-r_{3}}+\frac{1}{1-r_{4}}+...$. We can keenly set $y=\frac{1}{1-r}$ to get rid of the fractions and express it in a different variable. This leads us to $y(1-r)=1$ or $r=\frac{y-1}{y}$. Plugging it back to our original equation $(2[\frac{y-1}{y}])^{10}=1$, we get $\frac{2(y-1)^{10}}{(y-1)^{10}}=1$. By Vieta's, the sum of the roots is $\frac{-b}{a}$ which is equal to $\frac{m}{n}$. However, notice that $n$ is only needed, so we have to find the coefficient of the tenth-exponent. By the Binomial Theorem, this is equal to $2^{10}\binom{10}{0}y^{10}(-1)^0$ and by simplifying and factoring , we get $y^{10}(1024-1)=y^{10}(1023)$. Thus, $1023$ can be rewritten as $1023=(1024-1)=(2^{10}-1)=(2^5-1)(2^5+1)=31\cdot33=3\cdot11\cdot31$, and the final answer is $3+11+31=45$.