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

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.

2004 AMC 12B Problem 23

Answer: C

Difficulty rating: 2280

Solution:

Problem 23 in Other Years