AMC 10 · 2021 · #14

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 P(z) = z⁶ - 10z⁵ + Az⁴ + Bz³ + Cz² + Dz + 16 are POSITIVE INTEGERS (possibly repeated). Find B.

Pick an answer.

(A)

${-}88$

(B)

${-}80$

(C)

${-}64$

(D)

${-}41$

(E)

${-}40$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

By Vieta's formulas the roots sum to 10, multiply to 16, and B = -e₃, the negative of all triple-root products.

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

2STEP 2

Each root divides 16, and any root ≥ 4 pushes the sum past 10 — so every root is 1 or 2.

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

3STEP 3

Solving x + y = 6 and 2x + y = 10 gives four 2s and two 1s: the roots are {2, 2, 2, 2, 1, 1}.

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

4STEP 4

Count triple products by case. Three 2s: C(4, 3) triples each worth 8, contributing 32.

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

5STEP 5

Two 2s and a 1: 6·2 triples worth 4, contributing 48. One 2 and two 1s: 4 triples worth 2, contributing 8.

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

6STEP 6

The contributions add to e₃ = 32 + 48 + 8 = 88, so B = -e₃ = -88.

e₃ = 32 + 48 + 8 = 88, B = -88

7STEP 7

Among the choices -88, -80, -64, -41, -40, only -88 matches — answer (A).

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 10 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).

More like this