Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 9"
Hydroclippy (talk | contribs) |
|||
(15 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
− | + | ==Problem== | |
+ | ===Revised statement=== | ||
+ | Let <math>a_{n}</math> be a [[geometric sequence]] of [[complex number]]s with <math>a_{0}=1024</math> and <math>a_{10}=1</math>, and let <math>S</math> denote the [[infinite]] sum <math>S = a_{10}+a_{11}+a_{12}+...</math>. If the sum of all possible [[distinct]] values of <math>S</math> is <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are [[relatively prime]] [[positive integer]]s, compute the sum of the positive [[prime number | prime]] [[divisor | factors]] of <math>n</math>. | ||
− | [[Mock AIME 1 2006-2007]] | + | ===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]]. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <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> 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 Algebra Problems]] |
Latest revision as of 19:46, 20 October 2020
Contents
Problem
Revised statement
Let be a geometric sequence of complex numbers with
and
, and let
denote the infinite sum
. If the sum of all possible distinct values of
is
where
and
are relatively prime positive integers, compute the sum of the positive prime factors of
.
Original statement
Let be a geometric sequence for
with
and
. Let
denote the infinite sum:
. If the sum of all distinct values of
is
where
and
are relatively prime positive integers, then compute the sum of the positive prime factors of
.
Solutions
Solution 1
Let the ratio of consecutive terms of the sequence be . Then we have by the given that
so
and
, where
can be any of the tenth roots of unity.
Then the sum has value
. Different choices of
clearly lead to different values for
, so we don't need to worry about the distinctness condition in the problem. Then the value we want is
. Now, recall that if
are the
th roots of unity then for any integer
,
is 0 unless
in which case it is 1. Thus this simplifies to
where
.
Let ,
and
We seek , or the negative of the coefficient of
divided by the coefficient of
, which is
and
.
Therefore the answer is .
Solution 2
To answer the original problem statement, let the common geometric be and
denotes a term in the sequence. The term
can be represented as
and this expression has to be set equal to one. By simplifying, we get
where the polynomial
can have ten possible values or roots. Now, we take a look at the needed infinite geometric sum. The expression
can be expressed as
or even more better,
. However, the question is asking for the sum of the distinct roots of the polynomial, and as stated before,
is a tenth-degree polynomial and it had ten different roots. Therefore, our desired sum is
. We can keenly set
to get rid of the fractions and express it in a different variable. This leads us to
or
. Plugging it back to our original equation
, we get
. By Vieta's, the sum of the roots is
which is equal to
. However, notice that
is only needed, so we have to find the coefficient of the tenth-exponent. By the Binomial Theorem, this is equal to
and by simplifying and factoring , we get
. Thus,
can be rewritten as
, and the final answer is
.