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

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== Problem ==
 
== Problem ==
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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>
 +
 
 
== Solution ==
 
== Solution ==
−
{{solution}}
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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 formulas|Vieta's]] we know that <math>-z_4-z_3-z_2-z_1=-1</math>,
 +
so by [[Newton's 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{006}.</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
 +
 
 +
== Solution 3 ==
 +
 
 +
<math>P(x) = x^{2}Q(x)+x^{4}-x^{3}-x.</math>
 +
 
 +
So we just have to find: <math>\sum_{n=1}^{4} z^{4}_n - \sum_{n=1}^{4} z^{3}_n - \sum_{n=1}^{4} z_n</math>.
 +
 
 +
And by [[Newton's Sums]] this computes to: <math>11-4-1 = \boxed{006}</math>.
 +
 
 +
~ LuisFonseca123
 +
 
 +
== Solution 4 ==
 +
If we scale <math>Q(x)</math> by <math>x^2</math>, we get <math>x^6-x^5-x^4-x^2</math>. In order to get to <math>P(x)</math>, we add <math>x^4-x^3-x</math>. Therefore, our answer is <math>\sum_{n=1}^{4} z^4_n-z^3_n-z_n</math>. However, rearranging <math>Q(z_n) = 0</math>, makes our final answer <math>\sum_{n=1}^{4} z^2_n-z_n+1</math>. The sum of the squares of the roots is <math>1^2-2(-1) = 3</math> and the sum of the roots is <math>1</math>. Adding 4 to our sum, we get <math>3-1+4 = \boxed{006}</math>.
 +
 
 +
~ Vedoral
 +
 
 +
== Solution 5 ==
 +
Let <math>S_k</math> = <math>z_1^k+z_2^k+z_3^k+z_4^k</math>
 +
 
 +
By [[Newton's Sums]],
 +
 
 +
<math>S_1-1=0</math>
 +
 
 +
<math>S_2-S_1-2=0</math>
 +
 
 +
<math>S_3-S_2-S_1=0</math>
 +
 
 +
<math>S_4-S_3-S_2-4=0</math>
 +
 
 +
<math>S_5-S_4-S_3-S_1=0</math>
 +
 
 +
<math>S_6-S_5-S_4-S_2=0</math>
 +
 
 +
Solving for <math>S_1,S_2,S_3,S_4,S_5,S_6</math>, we get <math>S_1=1, S_2=3, S_3=4, S_4=11, S_5=16, S_6=30</math>
 +
 
 +
<math>P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{006}</math>
 +
 
 +
== Video Solution by Sal Khan ==
 +
https://www.youtube.com/watch?v=ZSESJ8TeGSI&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=14
 +
- AMBRIGGS
 +
 
 +
 
 +
[rule]
 +
 
 +
Nice!-sleepypuppy
  
 
== 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}}

Latest revision as of 01:41, 11 September 2024

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's 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 $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{006}.$

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

Solution 3

$P(x) = x^{2}Q(x)+x^{4}-x^{3}-x.$

So we just have to find: $\sum_{n=1}^{4} z^{4}_n - \sum_{n=1}^{4} z^{3}_n - \sum_{n=1}^{4} z_n$.

And by Newton's Sums this computes to: $11-4-1 = \boxed{006}$.

~ LuisFonseca123

Solution 4

If we scale $Q(x)$ by $x^2$, we get $x^6-x^5-x^4-x^2$. In order to get to $P(x)$, we add $x^4-x^3-x$. Therefore, our answer is $\sum_{n=1}^{4} z^4_n-z^3_n-z_n$. However, rearranging $Q(z_n) = 0$, makes our final answer $\sum_{n=1}^{4} z^2_n-z_n+1$. The sum of the squares of the roots is $1^2-2(-1) = 3$ and the sum of the roots is $1$. Adding 4 to our sum, we get $3-1+4 = \boxed{006}$.

~ Vedoral

Solution 5

Let $S_k$ = $z_1^k+z_2^k+z_3^k+z_4^k$

By Newton's Sums,

$S_1-1=0$

$S_2-S_1-2=0$

$S_3-S_2-S_1=0$

$S_4-S_3-S_2-4=0$

$S_5-S_4-S_3-S_1=0$

$S_6-S_5-S_4-S_2=0$

Solving for $S_1,S_2,S_3,S_4,S_5,S_6$, we get $S_1=1, S_2=3, S_3=4, S_4=11, S_5=16, S_6=30$

$P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{006}$

Video Solution by Sal Khan

https://www.youtube.com/watch?v=ZSESJ8TeGSI&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=14 - AMBRIGGS


[rule]

Nice!-sleepypuppy

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

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