2005 AIME II Problem 13

Attempt Problem 13 of the 2005 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AIME II solutions, or check the answer key.

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13.

Let P(x)P(x) be a polynomial with integer coefficients that satisfies P(17)=10P(17) = 10 and P(24)=17.P(24) = 17. Given that the equation P(n)=n+3P(n) = n + 3 has two distinct integer solutions n1n_1 and n2,n_2, find the product n1n2.n_1 \cdot n_2.

Answer: 418
Concepts:polynomialdivisibilityfactor
Difficulty rating: 2760
Solution:

Let S(x)=P(x)x3,S(x) = P(x) - x - 3, so S(17)=S(24)=10.S(17) = S(24) = -10. Since S(x)+10S(x) + 10 has integer coefficients and vanishes at 1717 and 24,24, S(x)=10+(x17)(x24)Q(x) \begin{aligned} S(x) &= -10 \\ &\quad {}+ (x - 17)(x - 24)Q(x) \end{aligned} for some polynomial QQ with integer coefficients.

If P(n)=n+3P(n) = n + 3 for an integer n,n, then (n17)(n24)Q(n)=10,(n-17)(n-24)Q(n) = 10, so (n17)(n24)(n-17)(n-24) divides 10.10. The factors n17n - 17 and n24n - 24 are integers differing by 77 whose product divides 10,10, so they are {2,5}\{2, -5\} or {5,2},\{5, -2\}, giving n=19n = 19 and n=22.n = 22. Both occur, for example, for P(x)=x7P(x) = x - 7 (x17)(x24).- (x-17)(x-24).

Hence n1n2=1922=418.n_1 \cdot n_2 = 19 \cdot 22 = 418.

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