Difference between revisions of "2003 AIME II Problems/Problem 9"

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== Problem ==
 
== Problem ==
An integer between 1000 and 9999, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?
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Consider the polynomials <math>P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x</math> and <math>Q(x) = x^{4} - x^{3} - x^{2} - 1.</math> Given that <math>z_{1},z_{2},z_{3},</math> and <math>z_{4}</math> are the roots of <math>Q(x) = 0,</math> find <math>P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).</math>
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== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}
  
 
== See also ==
 
== See also ==
* [[2003 AIME II Problems/Problem 8| Previous problem]]
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{{AIME box|year=2003|n=II|num-b=8|num-a=10}}
 
 
* [[2003 AIME II Problems/Problem 10| Next problem]]
 
 
 
* [[2003 AIME II Problems]]
 

Revision as of 13:39, 21 November 2007

Problem

Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$

Solution

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See also

2003 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AIME Problems and Solutions