AMC 10 · 2021 · #14
Grade 8 algebraProblem
All the roots of the polynomial z6−10z5+Az4+Bz3+Cz2+Dz+16 are positive integers, possibly repeated. What is the value of B?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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 $B$.
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: $-88,\;-80,\;-64,\;-41,\;-40$
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
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 $e_3 = \sum_{i<j<k} r_i r_j r_k$.
💡 Grade 8 polynomial identities: each coefficient is a symmetric function of the roots.
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 $2$.
💡 Grade 6 divisibility: roots are divisors of the product, and large divisors blow up the sum.
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: $2^4 \cdot 1^2 = 16$ ✓.
💡 Grade 8 system of linear equations in two unknowns.
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: $4 \cdot 8 = 32$.
💡 Grade 7 counting: pick how many of each kind, multiply by the product per case.
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.)
💡 Grade 7 product rule: independent choices multiply.
6.EE.A.3 Step 6 - Add the case contributions: $e_3 = 32 + 48 + 8 = 88$.
- Then $B = -e_3 = -88$.
💡 Grade 6 expressions: sum the case totals, then apply the Vieta sign.
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.
💡 Grade 6 multiple choice: match the computed coefficient to the list.
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^ 6.NS.B.4 Narrow the candidate roots. Since each $r_i$ is a positive integer divisor of $1 8.EE.C.8 Solve for how many $2$s and how many $1$s. Let $x$ = number of $2$s, $y$ = numbe 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 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.EE.A.3 Add the case contributions: $e_3 = 32 + 48 + 8 = 88$. Then $B = -e_3 = -88$. 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.3Apply 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.5Understand 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.4Find 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.8Find 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.1Know 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.8Analyze 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).
⭐ 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).
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