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

Solution 1

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

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 $19\cdot 22=\boxed{418}$ .

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 1

Solution 2

See also