2004 AMC 12B Problem 23

Question

The polynomial

x^3 - 2004x^2 + mx + n

has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of

n

are possible?

250{,}000

250{,}250

250{,}500

250{,}750

251{,}000

Answer: C

Concepts:Vieta’s Formulas polynomial counting integers in a range

Difficulty rating: 2280

Small hint:

The integer zero is the sum of the other two, which must be conjugates

\dfrac{a}{2} \pm r

Big hint:

The integer zero is half of

2004,

so it is

1002;

then count valid

r^2

Solution:

Let the integer zero be

a.

The other two zeros are irrational conjugates

\dfrac{a}{2} \pm r,

whose sum

a

equals the integer zero. Vieta’s formula on the

x^2

coefficient gives

a + a = 2004,

so

a = 1002

and the conjugate pair is

501 \pm r.

The coefficients are integers exactly when

r^2

is a positive integer, and the zeros are positive and distinct when

1 \le r^2 \le 501^2 - 1 = 251{,}000.

Since

r

cannot be an integer, we exclude the

500

perfect-square values

r^2 = 1^2, \ldots, 500^2,

leaving

251{,}000 - 500 = 250{,}500

values of

n.

Thus, the correct answer is C.