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2015 AMC 12A Problem 18

Below is the professionally curated solution for Problem 18 of the 2015 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:Vieta’s FormulasSimon’s Favorite Factoring Trick

Difficulty rating: 1990

18.

The zeros of the function f(x)=x2−ax+2af(x) = x^2 - ax + 2a are integers. What is the sum of the possible values of a?a?

77

88

1616

1717

1818

Solution:

Let the integer zeros be pp and q.q. By Vieta's formulas p+q=ap + q = a and pq=2a,pq = 2a, so pq=2(p+q).pq = 2(p+q). Rearranging gives (p−2)(q−2)=4.(p-2)(q-2) = 4.

The integer factor pairs of 44 are (1,4),(2,2),(4,1),(−1,−4),(−2,−2),(−4,−1),(1,4),(2,2),(4,1),(-1,-4),(-2,-2),(-4,-1), which yield (p,q)(p,q) pairs summing to a=9,8,−1,a = 9, 8, -1, and 0.0.

The distinct possible values of aa are 9,8,0,−1,9, 8, 0, -1, whose sum is 16.16.

Thus, the correct answer is C.

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