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 "2005 AIME II Problems/Problem 13"

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
We define <math>Q(x)=P(x)-x+7</math>, noting that it has roots at <math>17</math> and <math>24</math>. Hence <math>P(x)-x+7=A(x-17)(x-24)</math>. In particular, this means that  
 
We define <math>Q(x)=P(x)-x+7</math>, noting that it has roots at <math>17</math> and <math>24</math>. Hence <math>P(x)-x+7=A(x-17)(x-24)</math>. In particular, this means that  
 
<math>P(x)-x-3=A(x-17)(x-24)-10</math>. Therefore, <math>x=n_1,n_2</math> satisfy <math>A(x-17)(x-24)=10</math>, where <math>A</math>, <math>(x-17)</math>, and <math>(x-24)</math> are integers. This cannot occur if <math>x\le 17</math> or <math>x\ge 24</math> because the product <math>(x-17)(x-24)</math> will either be too large or not be a divisor of <math>10</math>. We find that <math>x=19</math> and <math>x=22</math> are the only values that allow <math>(x-17)(x-24)</math> to be a factor of <math>10</math>.  Hence the answer is <math>19\cdot 22=\boxed{418}</math>.
 
<math>P(x)-x-3=A(x-17)(x-24)-10</math>. Therefore, <math>x=n_1,n_2</math> satisfy <math>A(x-17)(x-24)=10</math>, where <math>A</math>, <math>(x-17)</math>, and <math>(x-24)</math> are integers. This cannot occur if <math>x\le 17</math> or <math>x\ge 24</math> because the product <math>(x-17)(x-24)</math> will either be too large or not be a divisor of <math>10</math>. We find that <math>x=19</math> and <math>x=22</math> are the only values that allow <math>(x-17)(x-24)</math> to be a factor of <math>10</math>.  Hence the answer is <math>19\cdot 22=\boxed{418}</math>.
 +
 +
 +
==Solution 2==
 +
 +
We know that <math>P(n)-(n+3)=0% so P(n) has two distinct solutions so %P(x)% is at least quadratic. Let us first try this problem out as if </math>P(x)<math> is a quadratic polynomial. Thus </math>P(x)=ax^2+bx+c<math> where </math>a,b,c$ are all integers.
  
 
== See also ==
 
== See also ==

Revision as of 22:30, 7 August 2018

Problem

Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$

Solution

We define $Q(x)=P(x)-x+7$, noting that it has roots at $17$ and $24$. Hence $P(x)-x+7=A(x-17)(x-24)$. In particular, this means that $P(x)-x-3=A(x-17)(x-24)-10$. Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$, where $A$, $(x-17)$, and $(x-24)$ are integers. This cannot occur if $x\le 17$ or $x\ge 24$ because the product $(x-17)(x-24)$ will either be too large or not be a divisor of $10$. We find that $x=19$ and $x=22$ are the only values that allow $(x-17)(x-24)$ to be a factor of $10$. Hence the answer is $19\cdot 22=\boxed{418}$.


Solution 2

We know that $P(n)-(n+3)=0% so P(n) has two distinct solutions so %P(x)% is at least quadratic. Let us first try this problem out as if$ (Error compiling LaTeX. Unknown error_msg)P(x)$is a quadratic polynomial. Thus$P(x)=ax^2+bx+c$where$a,b,c$ are all integers.

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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=2005_AIME_II_Problems/Problem_13&oldid=96856"