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 1)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
−
 
−
=== Solution 1 ===
 
−
 
−
Define the [[polynomial]] <math>Q(x) = P(x) - x + 7</math>.  By the givens, <math>Q(17) = 10 - 17 + 7 = 0</math>, <math>Q(24) = 17 - 24 + 7 = 0</math>, <math>Q(n_1) = n_1 + 3 - n_1 + 7 = 10</math> and <math>Q(n_2) = n_2 + 3 - n_2 + 7 = 10</math>.  Note that for any polynomial <math>P(x)</math> with [[integer]] [[coefficient]]s and any integers <math>a, b</math> we have <math>a - b</math> [[divisor|divides]] <math>P(a)-P(b)</math>.  So <math>n_1 - 17</math> divides <math>Q(n_1) - Q(17) = 10</math>, and so <math>n_1 - 17</math> must be one of the eight numbers <math>\pm1, \pm2, \pm5, \pm10</math> and so <math>n_1</math> must be one of the numbers <math>7, 12, 15, 16, 18, 19, 22</math> or <math>27</math>.  Similarly, <math>n_2 - 24</math> must divide <math>Q(n_2) - Q(24) = 10</math>, so <math>n_2</math> must be one of the eight numbers <math>14, 19, 22, 23, 25, 26, 29</math> or <math>34</math>.  Thus, <math>n_1</math> must be either 19 or 22.  Since <math>n_2</math> obeys the same conditions and <math>n_1</math> and <math>n_2</math> are different, one of them is <math>19</math> and the other is <math>22</math> and their product is <math>19 \cdot 22 = \boxed{418}</math>.
 
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 23:45, 14 February 2016

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

Solution 2

As above, 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 $19\cdot 22=\boxed{418}$.

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=75947"