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 10:09, 12 August 2008
Problem
Let
Let 
be the distinct zeros of 
and let 
for 
where 
and 
and 
are real numbers. Let
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 | ||
