Difference between revisions of "2013 AIME I Problems/Problem 10"

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==Problem 10==
 
==Problem 10==
 
There are nonzero integers <math>a</math>, <math>b</math>, <math>r</math>, and <math>s</math> such that the complex number <math>r+si</math> is a zero of the polynomial <math>P(x)={x}^{3}-a{x}^{2}+bx-65</math>. For each possible combination of <math>a</math> and <math>b</math>, let <math>{p}_{a,b}</math> be the sum of the zeros of <math>P(x)</math>. Find the sum of the <math>{p}_{a,b}</math>'s for all possible combinations of <math>a</math> and <math>b</math>.
 
There are nonzero integers <math>a</math>, <math>b</math>, <math>r</math>, and <math>s</math> such that the complex number <math>r+si</math> is a zero of the polynomial <math>P(x)={x}^{3}-a{x}^{2}+bx-65</math>. For each possible combination of <math>a</math> and <math>b</math>, let <math>{p}_{a,b}</math> be the sum of the zeros of <math>P(x)</math>. Find the sum of the <math>{p}_{a,b}</math>'s for all possible combinations of <math>a</math> and <math>b</math>.
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== Solution ==
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(solution)
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== See also ==
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{{AIME box|year=2013|n=I|num-b=9|num-a=11}}

Revision as of 20:45, 16 March 2013

Problem 10

There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$. For each possible combination of $a$ and $b$, let ${p}_{a,b}$ be the sum of the zeros of $P(x)$. Find the sum of the ${p}_{a,b}$'s for all possible combinations of $a$ and $b$.


Solution

(solution)

See also

2013 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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