AMC 10 · 2021 · #12

Grade 8 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

vieta-formulassystematic-enumerationpolynomial-factoringfactors systematic-enumerationcasework ↑ Prerequisites: vieta-formulas

📏 Long solution 💡 3 insights

Problem

All six roots of a degree-six polynomial are positive integers, possibly repeated. The roots add to 10 and multiply to 16. Find the coefficient of the cubic term.

Pick an answer.

(A)

${-}88$

(B)

${-}80$

(C)

${-}64$

(D)

${-}41$

(E)

${-}40$

How to solve

Strategy Systematic List

Tool #2 (Systematic List): the six positive-integer roots must sum to 10 and multiply to 16 = 2⁴. With those two strong constraints, the candidate multisets are very few — only powers of 2 (and 1s) can appear. Listing them systematically pins down the multiset {2, 2, 2, 2, 1, 1}. Tool #7 (Subproblems) breaks the work into: (a) use Vieta to relate B to a symmetric sum; (b) find the roots; (c) count triple products by case. Tool #6 (Guess and Check) checks each candidate multiset against the two constraints. Tool #3 (Eliminate) matches the final B = -88 to choice (A).

1STEP 1

Write the root relations

The target is a sum of triple products.

Σ r_i = 10, Π r_i = 16, B = - Σ₁ ≤ i ＜ j ＜ k ≤ 6 r_i r_j r_k

2STEP 2

Narrow the possible roots

Every root is a power of two.

r_i ∈ {1, 2, 4, 8, 16}, but r_i ≥ 4 → Σ r_i ＞ 10

3STEP 3

Fix how many of each

The sum and count give a small system.

x + y = 6 ; 2x + y = 10 → x = 4, y = 2

4STEP 4

Count the first kind of triple

Count triples of all the small root.

Case A: C(4, 3) · (2 · 2 · 2) = 4 · 8 = 32

5STEP 5

Count the mixed triples

Count the mixed triples too.

B: C(4, 2)C(2, 1) · 4 = 48, C: C(4, 1)C(2, 2) · 2 = 8

6STEP 6

Add them up

Add all the triple products.

e₃ = 32 + 48 + 8 = 88, B = -88

7STEP 7

Attach the sign

With the sign it is negative eighty-eight.

B = -88 → (A)

Answer

-88

Cross-check by expanding P(z) = (z-1)²(z-2)⁴. Mentally check the z⁵ coefficient: roots sum to 10, so coefficient of z⁵ is -10 ✓. Constant: (-1)⁶ · 1² · 2⁴ = 16 ✓. Quick partial expansion of (z-2)⁴ = z⁴ - 8z³ + 24z² - 32z + 16 and (z-1)² = z² - 2z + 1 — multiply and extract the z³ coefficient: from z² · (-8z³ · z⁰) contribution: hmm, easier — Vieta gave -88, which is between -100 (very rough upper magnitude C(6, 3) · max product) and 0. Sign and magnitude both reasonable; the careful casework above is the actual proof.

💡Key takeaway

This AMC 12 problem only needs Grade 8 systems of equations and basic counting you already know! The roots must sum to 10 and multiply to 16, so they can only be four 2s and two 1s. Count triple products by case (C(4, 3) · 8 = 32, C(4, 2)C(2, 1) · 4 = 48, C(4, 1)C(2, 2) · 2 = 8) — sum is 88, so B = -88, answer (A).