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Difference between revisions of "2005 AIME II Problems/Problem 13"

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== Solution ==
 
== Solution ==
  
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As above, 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  
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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>.
  

Revision as of 20:24, 13 October 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

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

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