Difference between revisions of "2003 AIME II Problems/Problem 15"
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Let | Let | ||
<center><math>P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).</math></center> | <center><math>P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).</math></center> | ||
− | Let <math>z_{1},z_{2},\ldots,z_{r}</math> be the distinct zeros of <math>P(x),</math> and let <math>z_{ | + | Let <math>z_{1},z_{2},\ldots,z_{r}</math> be the distinct zeros of <math>P(x),</math> and let <math>z_{k}^{2} = a_{k} + b_{k}i</math> for <math>k = 1,2,\ldots,r,</math> where <math>i = \sqrt { - 1},</math> and <math>a_{k}</math> and <math>b_{k}</math> are real numbers. Let |
<center><math>\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},</math></center> | <center><math>\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},</math></center> | ||
where <math>m,</math> <math>n,</math> and <math>p</math> are integers and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p.</math> | where <math>m,</math> <math>n,</math> and <math>p</math> are integers and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p.</math> |
Revision as of 11:09, 12 August 2008
Problem
Let
![$P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).$](http://latex.artofproblemsolving.com/a/d/b/adb0a8bb705a8514b88ca50f80f549d9059435eb.png)
Let be the distinct zeros of
and let
for
where
and
and
are real numbers. Let
![$\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},$](http://latex.artofproblemsolving.com/b/6/9/b6959c0d9b67d1a2d6914af2b95338ccf226924b.png)
where
and
are integers and
is not divisible by the square of any prime. Find
Solution
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See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |