Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (add problem)
m (→‎Solution)
(28 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
== 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?
+
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>
 +
 
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
When we use long division to divide <math>P(x)</math> by <math>Q(x)</math>, the remainder is <math>x^2-x+1</math>.
 +
 
 +
So, since <math>z_1</math> is a root, <math>P(z_1)=(z_1)^2-z_1+1</math>.
 +
 
 +
Now this also follows for all roots of <math>Q(x)</math>
 +
Now <cmath>P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1</cmath>
 +
 
 +
Now by Vieta's we know that <math>-z_4-z_3-z_2-z_1=-1</math>,
 +
so by Newton Sums we can find <math>z_1^2+z_2^2+z_3^2+z_4^2</math>
 +
 
 +
<math>a_ns_2+a_{n-1}s_1+2a_{n-2}=0</math>
 +
 
 +
<math>(1)(s_2)+(-1)(1)+2(-1)=0</math>
 +
 
 +
<math>s_2-1-2=0</math>
 +
 
 +
<math>s_2=3</math>
 +
 
 +
So finally
 +
<math>P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{6}.</math>
 +
 
 +
== Solution 2 ==
 +
 
 +
Let <math>S_k=z_1^k+z_2^k+z_3^k+z_4^k</math> then by [[Vieta's Formula]] we have
 +
<cmath>S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0</cmath>
 +
<cmath>S_0=4</cmath>
 +
<cmath>S_1=1</cmath>
 +
<cmath>S_2=3</cmath>
 +
By [[Newton's Sums]] we have
 +
<cmath>a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0</cmath>
 +
 
 +
Applying the formula couples of times yields <math>P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}</math>.
 +
 
 +
~ Nafer
  
 
== See also ==
 
== See also ==
−
* [[2003 AIME II Problems/Problem 8| Previous problem]]
+
{{AIME box|year=2003|n=II|num-b=8|num-a=10}}
−
 
 
−
* [[2003 AIME II Problems/Problem 10| Next problem]]
 
  
−
* [[2003 AIME II Problems]]
+
[[Category: Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 21:45, 2 July 2020

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

When we use long division to divide $P(x)$ by $Q(x)$, the remainder is $x^2-x+1$.

So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$.

Now this also follows for all roots of $Q(x)$ Now \[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\]

Now by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$, so by Newton Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$

$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$

$(1)(s_2)+(-1)(1)+2(-1)=0$

$s_2-1-2=0$

$s_2=3$

So finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{6}.$

Solution 2

Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0=4\] \[S_1=1\] \[S_2=3\] By Newton's Sums we have \[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\]

Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}$.

~ Nafer

See also

2003 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_9&oldid=127331"