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2005 AIME II Problems/Problem 13

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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.

See also

2005 AIME II (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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