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

(Solution)
(Solution 2)
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==Solution 2==
 
==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.
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We know that <math>P(n)-(n+3)=0</math> so P(n) has two distinct solutions so <math>P(x)</math> 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</math> are all integers.
  
 
== See also ==
 
== See also ==

Revision as of 23:31, 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 $P(x)$ is a quadratic polynomial. Thus $P(x)=ax^2+bx+c$ where $a,b,c$ are all integers.

See also

2005 AIME II (ProblemsAnswer KeyResources)
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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