2015 AMC 12A Problem 18

Attempt Problem 18 of the 2015 AMC 12A 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 2015 AMC 12A solutions, or check the answer key.

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18.

The zeros of the function

f(x) = x^2 - ax + 2a

are integers. What is the sum of the possible values of

a?

7

8

16

17

18

If the integer roots are

p,q

then

p+q = a

and

pq = 2a

Eliminate

a

to get

pq = 2(p+q),

i.e.

(p-2)(q-2) = 4

Let the integer zeros be

p

and

q.

By Vieta’s formulas

p + q = a

and

pq = 2a,

pq = 2(p+q).

Rearranging gives

(p-2)(q-2) = 4.

The integer factor pairs of

4

are

(1,4),

(2,2),

(4,1),

(-1,-4),

(-2,-2),

(-4,-1),

which yield

(p,q)

pairs summing to

a = 9, 8, -1,

and

0.

The distinct possible values of

a

are

9, 8, 0, -1,

whose sum is

16.

Thus, the correct answer is C.

Problem 18 in other years