AMC 10 · 2004 · #23

Grade 11 algebra
vieta-formulaspolynomial-rootsperfect-squares convert-to-algebracomplementary-counting ↑ Prerequisites: vieta-formulas
📏 Long solution 💡 4 insights
Problem
A cubic with integer coefficients has three different positive zeros, exactly one of them an integer. That integer zero equals the sum of the other two. Count how many different values the constant term can take.

Pick an answer.

(A)
$250,\!000$
(B)
$250,\!250$
(C)
$250,\!500$
(D)
$250,\!750$
(E)
$251,\!000$
How to solve
Strategy Introduce a Variable

The zeros carry all the information, so name them first and let Vieta's formulas convert the coefficient -2004 into a statement about the zeros. That collapses the whole family of polynomials down to a single unknown number, the product of the two non-integer zeros. Counting values of n then becomes counting allowed values of that one integer, which is an interval of integers minus a short list of forbidden ones — a job for the complement.

1STEP 1

Name the three zeros

Naming the hidden zeros makes the third one their sum.

x³ - 2004x² + mx + n = (x - r) (x - s) (x - (r+s))
2STEP 2

Read off the sum of the zeros

The leading coefficients fix that integer zero at 1002.

2(r+s) = 2004 ⟹ r + s = 1002
3STEP 3

Reduce everything to one number

A single product then controls both unknown coefficients.

m = t + 1002², n = -1002 t, t = rs ∈ Z
4STEP 4

Write the two hidden zeros explicitly

The quadratic formula writes both hidden zeros using that one number.

r, s = (1002 ± √(1002² - 4t))/2 = 501 ± √(251001 - t)
5STEP 5

Find the range of t

Real, distinct and positive squeeze it to 251000 candidates.

0 < t < 251001, t ∈ Z ⟹ t ∈ {1, 2, …, 251000}
6STEP 6

Spot which t are forbidden

The forbidden ones are exactly those making a perfect square.

r, s ∉ Z ⇔ 251001 - t is not a perfect square
7STEP 7

Count the forbidden ones

Counting those squares gives exactly 500 to remove.

500² = 250000 ≤ 251000 < 251001 = 501² ⟹ k ∈ {1, …, 500}
8STEP 8

Subtract and translate back to n

Subtracting leaves 250,500 values, choice (C).

251000 - 500 = 250500
Answer
250, 500
The count factors as 250500 = 501 · 500, and that is not an accident: writing the sum of the two hidden zeros as 2S (here 2S = 1002, so S = 501), the argument gives t ∈ {1, …, S² - 1} minus the S - 1 values with S² - t a perfect square, that is (S² - 1) - (S - 1) = S(S-1). Testing that formula on a tiny version confirms it. Replace 2004 by 6, so S = 3: the zeros are 3 and 3 ± √(9 - t), and t runs over 1 through 8. Listing them, t = 1, 2, 3, 4, 6, 7 give irrational pairs such as 3 ± √(8), while t = 5 gives 1 and 5 and t = 8 gives 2 and 4 — both integer pairs, both rejected. That leaves 6 = 3 · 2 = S(S-1) values, matching the formula. Applying it to S = 501 gives 501 · 500 = 250500, which is choice (C). The nearby choices are also sensible-looking multiples of 250, so the exact bookkeeping — including the fact that 501² = 251001 just misses the range — is what separates them.
💡Key takeaway

When a polynomial's zeros are tied together by a rule, describe the whole family with one number, then count which values of that number obey every rule.

  • Name the three zeros
  • Read off the sum of the zeros
  • Reduce everything to one number
  • Write the two hidden zeros explicitly
  • Find the range of t
  • Spot which t are forbidden
  • Count the forbidden ones
  • Subtract and translate back to n