AMC 10 · 2021 · #14

Grade 8 algebra

Archetype Small-Domain Constrained Search

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

📏 Long solution 💡 3 insights

Problem

All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$ ?

$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$

Pick an answer.

(A)

${-}88$

(B)

${-}80$

(C)

${-}64$

(D)

${-}41$

(E)

${-}40$

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Toolkit + CCSS Solution

Understand

Restated: All six roots of $P(z) = z^6 - 10z^5 + Az^4 + Bz^3 + Cz^2 + Dz + 16$ are POSITIVE INTEGERS (possibly repeated). Find $B$.

Givens: Polynomial: $z^6 - 10z^5 + Az^4 + Bz^3 + Cz^2 + Dz + 16$, leading coefficient $1$; All six roots are positive integers, repeats allowed; Coefficient of $z^5$ is $-10$ (so sum of roots is $10$); Constant term is $16$ (so product of roots is $16$); Answer choices: $-88,\;-80,\;-64,\;-41,\;-40$

Unknowns: The coefficient $B$ (coefficient of $z^3$)

Understand

Restated: All six roots of $P(z) = z^6 - 10z^5 + Az^4 + Bz^3 + Cz^2 + Dz + 16$ are POSITIVE INTEGERS (possibly repeated). Find $B$.

Plan

Primary tool: #2 Systematic List

Secondary: #7 Identify Subproblems, #6 Guess and Check, #3 Eliminate Possibilities

Tool #2 (Systematic List): the six positive-integer roots must sum to $10$ and multiply to $16 = 2^4$. With those two strong constraints, the candidate multisets are very few — only powers of $2$ (and $1$s) 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).

Execute — Answer: A

#7 Identify Subproblems 8.EE.A.1 Step 1

Apply Vieta's formulas.
For a monic polynomial $z^6 + c_5 z^5 + c_4 z^4 + c_3 z^3 + \cdots$ with roots $r_1, \dots, r_6$: the coefficient of $z^5$ equals $-(r_1 + \cdots + r_6)$, the constant term equals $r_1 r_2 \cdots r_6$, and the coefficient of $z^3$ equals $-(\text{sum of products of roots taken three at a time})$.
Reading off from our polynomial: sum $= 10$, product $= 16$, and $B = -e_3$ where $e_3 = \sum_{i<j<k} r_i r_j r_k$.

$$\sum r_i = 10,\quad \prod r_i = 16,\quad B = -\!\!\!\sum_{1\le i<j<k\le 6}\!\!\! r_i r_j r_k$$

💡 Grade 8 polynomial identities: each coefficient is a symmetric function of the roots.

#6 Guess and Check 6.NS.B.4 Step 2

Narrow the candidate roots.
Since each $r_i$ is a positive integer divisor of $16$, the only options are $1, 2, 4, 8, 16$.
But any root $\ge 4$ forces the sum to exceed $10$ (e.g.
one root $4$ plus five roots $\ge 1$ gives sum $\ge 9$; matching product $16$ then needs the other five to multiply to $4$, but that minimum sum is $4+1+1+1+1 = 8$ pushing total to $\ge 12 > 10$, contradiction).
So every root is $1$ or $2$.

$$r_i \in \{1, 2, 4, 8, 16\},\;\; \text{but } r_i \ge 4 \Rightarrow \sum r_i > 10$$

💡 Grade 6 divisibility: roots are divisors of the product, and large divisors blow up the sum.

#2 Systematic List 8.EE.C.8 Step 3

Solve for how many $2$s and how many $1$s.
Let $x$ = number of $2$s, $y$ = number of $1$s.
Then $x + y = 6$ and $2x + y = 10$.
Subtracting: $x = 4$, so $y = 2$.
The multiset of roots is $\{2, 2, 2, 2, 1, 1\}$.
Verify the product: $2^4 \cdot 1^2 = 16$ ✓.

$$\begin{cases} x + y = 6 \\ 2x + y = 10 \end{cases} \Rightarrow x = 4,\; y = 2$$

💡 Grade 8 system of linear equations in two unknowns.

#2 Systematic List 7.SP.C.8 Step 4

Compute $e_3 = \sum r_i r_j r_k$ by casework on how many $2$s appear in the chosen triple.
(Case A) Three $2$s: choose $3$ of the four $2$s, $\binom{4}{3} = 4$ ways, each triple's product is $8$.
Contribution: $4 \cdot 8 = 32$.

$$\text{Case A: } \binom{4}{3} \cdot (2\cdot 2\cdot 2) = 4 \cdot 8 = 32$$

💡 Grade 7 counting: pick how many of each kind, multiply by the product per case.

#2 Systematic List 7.SP.C.8 Step 5

(Case B) Two $2$s and one $1$: $\binom{4}{2}$ for the $2$s and $\binom{2}{1}$ for the $1$, product $4$.
Contribution: $6 \cdot 2 \cdot 4 = 48$.
(Case C) One $2$ and two $1$s: $\binom{4}{1} \cdot \binom{2}{2}$, product $2$.
Contribution: $4 \cdot 1 \cdot 2 = 8$.
(No case D: there are only two $1$s, so we can't pick three.)

$$\text{B: } \binom{4}{2}\binom{2}{1} \cdot 4 = 48,\;\; \text{C: } \binom{4}{1}\binom{2}{2} \cdot 2 = 8$$

💡 Grade 7 product rule: independent choices multiply.

#7 Identify Subproblems 6.EE.A.3 Step 6

Add the case contributions: $e_3 = 32 + 48 + 8 = 88$.
Then $B = -e_3 = -88$.

$$e_3 = 32 + 48 + 8 = 88,\quad B = -88$$

💡 Grade 6 expressions: sum the case totals, then apply the Vieta sign.

#3 Eliminate Possibilities 6.EE.B.5 Step 7

Match $-88$ to the answer choices.
It's (A).
The other values ($-80, -64, -41, -40$) don't agree with the case totals.

$$B = -88 \;\Rightarrow\; \textbf{(A)}$$

💡 Grade 6 multiple choice: match the computed coefficient to the list.

[1] #7 8.EE.A.1 Apply Vieta's formulas. For a monic polynomial $z^6 + c_5 z^5 + c_4 z^4 + c_3 z^

[2] #6 6.NS.B.4 Narrow the candidate roots. Since each $r_i$ is a positive integer divisor of $1

[3] #2 8.EE.C.8 Solve for how many $2$s and how many $1$s. Let $x$ = number of $2$s, $y$ = numbe

[4] #2 7.SP.C.8 Compute $e_3 = \sum r_i r_j r_k$ by casework on how many $2$s appear in the chos

[5] #2 7.SP.C.8 (Case B) Two $2$s and one $1$: $\binom{4}{2}$ for the $2$s and $\binom{2}{1}$ fo

[6] #7 6.EE.A.3 Add the case contributions: $e_3 = 32 + 48 + 8 = 88$. Then $B = -e_3 = -88$.

[7] #3 6.EE.B.5 Match $-88$ to the answer choices. It's (A). The other values ($-80, -64, -41, -

Review

Reasonableness: Cross-check by expanding $P(z) = (z-1)^2(z-2)^4$. Mentally check the $z^5$ coefficient: roots sum to $10$, so coefficient of $z^5$ is $-10$ ✓. Constant: $(-1)^6 \cdot 1^2 \cdot 2^4 = 16$ ✓. Quick partial expansion of $(z-2)^4 = z^4 - 8z^3 + 24z^2 - 32z + 16$ and $(z-1)^2 = z^2 - 2z + 1$ — multiply and extract the $z^3$ coefficient: from $z^2 \cdot (-8z^3 \cdot z^0)$ contribution: hmm, easier — Vieta gave $-88$, which is between $-100$ (very rough upper magnitude $\binom{6}{3} \cdot \text{max product}$) and $0$. Sign and magnitude both reasonable; the careful casework above is the actual proof.

Alternative: Tool #6 (Guess and Check) by direct expansion: write $P(z) = (z-1)^2 (z-2)^4 = (z^2 - 2z + 1)(z^4 - 8z^3 + 24z^2 - 32z + 16)$ and multiply. The $z^3$ terms come from $z^2 \cdot (-32z) = -32 z^3$, $(-2z) \cdot 24 z^2 = -48 z^3$, $1 \cdot (-8z^3) = -8 z^3$. Sum: $-32 - 48 - 8 = -88$. Same answer (A).

CCSS standards used (min grade 8)

6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Combining the three case contributions $32 + 48 + 8 = 88$ and applying the Vieta sign.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Matching the computed $B = -88$ against the five answer choices to select (A).)
6.NS.B.4 Find the greatest common factor of two whole numbers (Using divisibility ($r_i$ must divide $16$) to narrow root candidates to $\{1, 2\}$.)
7.SP.C.8 Find probabilities of compound events including organized lists, tables, tree diagrams (Counting triples by case using $\binom{4}{3}, \binom{4}{2}\binom{2}{1}, \binom{4}{1}\binom{2}{2}$.)
8.EE.A.1 Know and apply the properties of integer exponents (Stating Vieta's formulas relating the polynomial coefficients to symmetric sums of the six roots.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Solving $x + y = 6$ and $2x + y = 10$ to find four $2$s and two $1$s.)

⭐ 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 $2$s and two $1$s. Count triple products by case ($\binom{4}{3} \cdot 8 = 32$, $\binom{4}{2}\binom{2}{1} \cdot 4 = 48$, $\binom{4}{1}\binom{2}{2} \cdot 2 = 8$) — sum is $88$, so $B = -88$, answer (A).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

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