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

Revision as of 17:20, 22 March 2007

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

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 $R(x)$ 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, $n_1 - 24$ must divide $Q(n_1) - Q(24) = 10$ , so $n_1$ 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 $19 \cdot 22 = 418$ .

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 == See also ==
+{{AIME box|year=2005|n=II|num-b=12|num-a=14}}
-* [[2005 AIME II Problems/Problem 12| Previous problem]]
-* [[2005 AIME II Problems/Problem 14| Next problem]]
-* [[2005 AIME II Problems]]
 [[Category:Intermediate Algebra Problems]]

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

Revision as of 17:20, 22 March 2007

Problem

Solution

See also