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

Latest revision as of 18:46, 20 October 2020

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 $3+11+31=45$ .

@@ Line 4: / Line 4: @@
 ===Original statement===
+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>.
+==Solutions==
+===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]].
-==Solution==
+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 [[roots 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
-{{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>,
-*[[Mock AIME 1 2006-2007/Problem 8 | Previous Problem]]
+and <math>2^{10}(1-1/t)^{10}-1=0\implies2^{10}(t-1)^{10}-t^{10}=0</math>
-*[[Mock AIME 1 2006-2007/Problem 10 | Next Problem]]
+We seek <math>\sum t</math>, or the negative of the coefficient of <math>t^9</math> divided by the coefficient of <math>t^10</math>, which is <math>2^{10}\cdot10/(2^{10}-1)=2^{11}\cdot5/(2^{10}-1)</math> and <math>2^{10}-1=33*31=3*11*31</math>.
+Therefore the answer is <math>\boxed{45}</math>.
+===Solution 2===
+To answer the original problem statement, let the common geometric be <math>r</math> and <math>a_n</math> denotes a term in the sequence. The term <math>a_{10}</math> can be represented as <math>a\cdot r^{10}</math> and this expression has to be set equal to one. By simplifying, we get <math>(2r)^{10}=1</math> where the polynomial <math>r^{10}</math> can have ten possible values or roots. Now, we take a look at the needed infinite geometric sum. The expression <math>a_{10}+a_{11}+a_{12}+...</math> can be expressed as <math>1+r+r^2+r^3+...</math> or even more better, <math>\frac{1}{1-r}</math>. However, the question is asking for the sum of the distinct roots of the polynomial, and as stated before, <math>r^{10}</math> is a tenth-degree polynomial and it had ten different roots. Therefore, our desired sum is <math>\frac{1}{1-r_{1}}+\frac{1}{1-r_{2}}+\frac{1}{1-r_{3}}+\frac{1}{1-r_{4}}+...</math>. We can keenly set <math>y=\frac{1}{1-r}</math> to get rid of the fractions and express it in a different variable. This leads us to <math>y(1-r)=1</math> or <math>r=\frac{y-1}{y}</math>. Plugging it back to our original equation <math>(2[\frac{y-1}{y}])^{10}=1</math>, we get <math>\frac{2(y-1)^{10}}{(y-1)^{10}}=1</math>. By Vieta's, the sum of the roots is <math>\frac{-b}{a}</math> which is equal to <math>\frac{m}{n}</math>. However, notice that <math>n</math> is only needed, so we have to find the coefficient of the tenth-exponent. By the Binomial Theorem, this is equal to <math>2^{10}\binom{10}{0}y^{10}(-1)^0</math> and by simplifying and factoring , we get <math>y^{10}(1024-1)=y^{10}(1023)</math>. Thus, <math>1023</math> can be rewritten as <math>1023=(1024-1)=(2^{10}-1)=(2^5-1)(2^5+1)=31\cdot33=3\cdot11\cdot31</math>, and the final answer is <math>3+11+31=45</math>.
+==See Also==
+*[[Mock AIME 1 2006-2007 Problems/Problem 8 | Previous Problem]]
+*[[Mock AIME 1 2006-2007 Problems/Problem 10 | Next Problem]]
 *[[Mock AIME 1 2006-2007]]
-[[Category:Intermediate Complex Numbers Problems]]
+[[Category:Intermediate Algebra Problems]]

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