Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 9"
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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> 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 | + | Therefore the answer is 45. |
Revision as of 19:51, 25 January 2010
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
divided by the coefficient of
, which is
and
.
Therefore the answer is 45.