1987 AMC 12 Problem 23

Attempt Problem 23 of the 1987 AMC 12 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 1987 AMC 12 solutions, or check the answer key.

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23.

p

is a prime and both roots of

x^2+px-444p=0

are integers, then

1\lt p\le11

11\lt p\le21

21\lt p\le31

31\lt p\le41

41\lt p\le51

For an integer root

x,

rewrite the equation as

x^2=p(444-x)

Since

p

is prime, write

x=np

and factor

444

An integer root satisfies

x^2=p(444-x).

Thus

x

is divisible by

p,

so write

x=np.

Substitution gives

n(n+1)p=444=2^2\cdot3\cdot37.

The consecutive product condition works only with

p=37,

for which

n=3

-4.

The roots are

111

and

-148,

31\lt p\le41.

Thus the correct answer is D.

Problem 23 in other years