Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 9"
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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 ... | 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: | Another solution: | ||
− | <math>\sum\frac1{1-z/2}</math> where <math>z^10=1</math>. | + | <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>, | 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> | + | 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>. | + | 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 | Therefore the answer is 2+5=7 |
Revision as of 01:06, 5 November 2009
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
.
Solution
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 ...
Another solution:
where
.
Let ,
and
We seek , or the negative of the coefficient of
, which is
.
Therefore the answer is 2+5=7