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 "2010 AIME II Problems/Problem 6"

m (→‎Solution)
(Semi-automated contest formatting - script by azjps)
Line 1: Line 1:
−
==Problem==
+
== Problem ==
−
 
+
Find the smallest positive integer <math>n</math> with the property that the [[polynomial]] <math>x^4 - nx + 63</math> can be written as a product of two nonconstant polynomials with integer coefficients.  
−
Find the smallest positive integer n with the property that the polynomial <math>x^4 - nx + 63</math> can be written as a product of two nonconstant polynomials with integer coefficients.  
 
  
 
==Solution==
 
==Solution==
−
 
+
There are two ways for a monic fourth degree polynomial to be factored into two non-constant polynomials with real coefficients: into a [[cubic]] and a linear equation, or 2 [[quadratic]]s.
−
There are 2 ways for a monic fourth degree polynomial to be factored, into a cubic and a linear equation, or 2 quadratics.
 
  
 
<br/>
 
<br/>
−
<b>Case 1)</b> cubic and linear
+
*'''Case 1''': The factors are cubic and linear.
−
 
 
−
Let <math>(x-r_1)</math> be the linear equation (it must contain one root of the quatic)
 
−
 
 
−
and <math>x^3+ax^2+bx+c</math> be the cubic.  
 
  
−
By rational roots theorem, <math>r_1=1,3,7, 9</math>, or <math>63</math>
+
Let <math>x-r_1</math> be the linear root, where <math>r_1</math> is a root of the given quartic, and let <math>x^3+ax^2+bx+c</math> be the cubic.
  
−
<math>(x^3+ax^2+bx+c)(x-r_1)=x^4+(a-r_1)x^3+(b-ar_1)x^2+(c-br_1)x-cr_1</math>
+
By the [[rational roots theorem]], then <math>r_1=1,3,7, 9</math>, or <math>63</math>. Observe that
  
−
So <math>a-r_1=0 \rightarrow a=r_1</math>, <math>b-ar_1=0\rightarrow b=a^2</math>,
+
<cmath>(x^3+ax^2+bx+c)(x-r_1)=x^4+(a-r_1)x^3+(b-ar_1)x^2+(c-br_1)x-cr_1.</cmath>
  
−
<math>c-br_1=-n\rightarrow n=a^3-c</math>, and <math>-cr_1=63 \rightarrow c=\frac{-63}{a}</math>.
+
Setting [[coefficient]]s equal, we have <math>a-r_1=0 \Longrightarrow a=r_1</math>, <math>b-ar_1=0\Longrightarrow b=a^2</math>, and <math>c-br_1=-n \Longrightarrow n=a^3-c</math>, and <math>-cr_1=63 \Longrightarrow c=\frac{-63}{a}</math>.
  
 
<br/>
 
<br/>
  
−
So <math>n=a^3+\frac{63}{a}</math>, <math>r_1=1,3,7, 9</math>, or <math>63</math>, which reach minimum when <math>r_1=3</math>, where <math>n=48</math>
+
It follows that <math>n=a^3+\frac{63}{a}</math>, <math>r_1=1,3,7, 9</math>, or <math>63</math>, which reach minimum when <math>r_1=3</math>, where <math>n=48</math>.
  
−
<br/>
 
  
 
<br/>
 
<br/>
−
<b>Case 2)</b> 2 quadratic
+
*'''Case 2''': The factors are quadratics.
−
 
 
−
Let <math>x^2+ax+b</math> and <math>x^2+cx+d</math> be the two quadratics,
 
−
 
 
−
<math>(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd</math>
 
  
−
Therefore, we have
+
Let <math>x^2+ax+b</math> and <math>x^2+cx+d</math> be the two quadratics, so that
  
−
<math>a + c = 0\rightarrow a=-c</math>, <math>b + d + ac = 0\rightarrow b+d=a^2</math> , <math>ad + bc = - n</math>
+
<cmath>(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.</cmath>
  
−
and <math>bd = 63</math>.
+
Therefore, again setting coefficients equal, <math>a + c = 0\Longrightarrow a=-c</math>, <math>b + d + ac = 0\Longrightarrow b+d=a^2</math> , <math>ad + bc = - n</math>, and so <math>bd = 63</math>.
  
−
<math>b+d=a^2</math> ,hence the only possible values for (b,d) are (1,63) and (7,9). From this we find that the possible values for n are <math>\pm 8 * 62</math> and <math>\pm 4 * 2</math>. Therefore, the answer is <math>\boxed{8}</math>.
+
Since <math>b+d=a^2</math>, the only possible values for <math>(b,d)</math> are <math>(1,63)</math> and <math>(7,9)</math>. From this we find that the possible values for <math>n</math> are <math>\pm 8 \cdot 62</math> and <math>\pm 4 \cdot 2</math>. Therefore, the answer is <math>\boxed{008}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=5|num-a=7|n=II}}
 
{{AIME box|year=2010|num-b=5|num-a=7|n=II}}
 +
 +
[[Category:Intermediate Algebra Problems]]

Revision as of 10:45, 6 April 2010

Problem

Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients.

Solution

There are two ways for a monic fourth degree polynomial to be factored into two non-constant polynomials with real coefficients: into a cubic and a linear equation, or 2 quadratics.


  • Case 1: The factors are cubic and linear.

Let $x-r_1$ be the linear root, where $r_1$ is a root of the given quartic, and let $x^3+ax^2+bx+c$ be the cubic.

By the rational roots theorem, then $r_1=1,3,7, 9$, or $63$. Observe that

\[(x^3+ax^2+bx+c)(x-r_1)=x^4+(a-r_1)x^3+(b-ar_1)x^2+(c-br_1)x-cr_1.\]

Setting coefficients equal, we have $a-r_1=0 \Longrightarrow a=r_1$, $b-ar_1=0\Longrightarrow b=a^2$, and $c-br_1=-n \Longrightarrow n=a^3-c$, and $-cr_1=63 \Longrightarrow c=\frac{-63}{a}$.


It follows that $n=a^3+\frac{63}{a}$, $r_1=1,3,7, 9$, or $63$, which reach minimum when $r_1=3$, where $n=48$.



  • Case 2: The factors are quadratics.

Let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that

\[(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.\]

Therefore, again setting coefficients equal, $a + c = 0\Longrightarrow a=-c$, $b + d + ac = 0\Longrightarrow b+d=a^2$ , $ad + bc = - n$, and so $bd = 63$.

Since $b+d=a^2$, the only possible values for $(b,d)$ are $(1,63)$ and $(7,9)$. From this we find that the possible values for $n$ are $\pm 8 \cdot 62$ and $\pm 4 \cdot 2$. Therefore, the answer is $\boxed{008}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_6&oldid=34233"