AMC 10 · 2005 · #24

Grade 11 algebra

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

polynomial-rootspolynomial-factoringfunction-compositionfundamental-counting-principle identify-subproblemscomplementary-countingsystematic-enumeration ↑ Prerequisites: polynomial-rootsfunction-compositionpolynomial-factoring

📏 Long solution 💡 4 insights

Problem

A cubic with three distinct roots is composed with an unknown polynomial. The result must equal the cubic times a polynomial of degree exactly three. Count the unknown polynomials that work.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Identify Subproblems

The condition is an identity between polynomials, which is hard to attack head on, so break it into two independent pieces: what degree Q must have, and what values Q must take at 1, 2, 3. Comparing degrees settles the first piece instantly. Rewriting the identity as a divisibility statement and using the distinct roots of P settles the second. What is left is a small list of value triples, and the ones that fail are easier to describe than the ones that work, so count the complement.

1STEP 1

Pin down the degree of Q

Comparing degrees pins the unknown at degree two.

deg(P(Q(x))) = 3deg Q = deg(P(x)R(x)) = 3 + 3 = 6 ⟹ deg Q = 2

2STEP 2

Rewrite the identity as divisibility

The identity is really a divisibility statement.

P(x) ∣ F(x) ⇔ F(1) = F(2) = F(3) = 0

3STEP 3

Turn the roots into conditions on Q

Distinct roots turn that into three simple value conditions.

P(Q(k)) = (Q(k)-1)(Q(k)-2)(Q(k)-3) = 0 ⇔ Q(k) ∈ {1, 2, 3}

4STEP 4

Check those conditions are enough

Those conditions are also sufficient, which makes the count exact.

R(x) = P(Q(x))/P(x), deg R = 6 - 3 = 3

5STEP 5

Count candidate value triples

Three values at three inputs determine the polynomial, giving 27 candidates.

3 · 3 · 3 = 27

6STEP 6

Spot the ones that flatten out

A candidate flattens exactly when the three values form an arithmetic progression.

Q(1) - 2Q(2) + Q(3) = (p+q+r) - 2(4p+2q+r) + (9p+3q+r) = 2p

7STEP 7

List the failures

Listing those gives 5 failures.

(1,1,1), (2,2,2), (3,3,3), (1,2,3), (3,2,1)

8STEP 8

Subtract

Subtracting leaves 22, choice (B).

27 - 5 = 22

Answer

The wrong answer choices are exactly the places a count can stop too early: 27 forgets that Q must have degree exactly 2, and 24 removes only the three constant maps while keeping Q(x) = x and Q(x) = 4-x. A spot check on a surviving triple confirms the machinery: (Q(1), Q(2), Q(3)) = (1,1,2) gives p = (1 - 2 + 2)/2 = 1/2 and then Q(x) = 1/2x² - 3/2x + 2, which is genuinely quadratic, does send 1, 2, 3 to 1, 1, 2, and makes P(Q(x)) vanish at x = 1, 2, 3 as required. Note also that the degree of R never rules anything out at the end: as soon as deg Q = 2, the quotient automatically has degree 3.

💡Key takeaway

A polynomial identity usually collapses to a handful of conditions at the roots; find those conditions, then it is just counting.

Pin down the degree of Q
Rewrite the identity as divisibility
Turn the roots into conditions on Q
Check those conditions are enough
Count candidate value triples
Spot the ones that flatten out
List the failures
Subtract