2005 AIME II Problems/Problem 13

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.

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2005 AIME II Problems/Problem 13

Contents

Problem

Solution

Solution 2

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