AMC 10 · 2010 · #21
Grade 7 algebraThe polynomial x3−ax2+bx−2010 has three positive integer roots. What is the smallest possible value of a?
Pick an answer.
AMC 10 2010 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: The cubic $x^3-ax^2+bx-2010$ has three roots, and all three are positive whole numbers. Among every cubic of this shape that works, find the smallest possible value of the coefficient $a$.
Givens: The polynomial is $x^3-ax^2+bx-2010$ with leading coefficient $1$; Its three roots are all positive integers; The constant term is $-2010$; Answer choices: (A) $78$, (B) $88$, (C) $98$, (D) $108$, (E) $118$
Unknowns: The smallest possible value of $a$
Understand
Restated: The cubic $x^3-ax^2+bx-2010$ has three roots, and all three are positive whole numbers. Among every cubic of this shape that works, find the smallest possible value of the coefficient $a$.
Givens: The polynomial is $x^3-ax^2+bx-2010$ with leading coefficient $1$; Its three roots are all positive integers; The constant term is $-2010$; Answer choices: (A) $78$, (B) $88$, (C) $98$, (D) $108$, (E) $118$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #2 Make a Systematic List
Tool #14 (Extreme Principle): the question asks for the smallest $a$, so once we know what $a$ depends on we push that quantity to its minimum. Tool #4 (Introduce a Variable): name the three roots $p,q,r$ so the polynomial becomes a product of linear factors we can read coefficients from. Tool #7 (Identify Subproblems): split the work into (i) link $a$ to the roots, (ii) factor $2010$, (iii) group the factors, (iv) minimize. Tool #2 (Make a Systematic List): there are only a few ways to split $2010$ into three whole-number factors, so list them all and compare their sums.
Execute — Answer: A
6.EE.A.2 Step 1 Name the roots and factor the cubic
- Call the three positive integer roots $p$, $q$, and $r$.
- A cubic whose leading coefficient is $1$ and whose roots are $p,q,r$ can be written as the product $(x-p)(x-q)(x-r)$.
- This is the same polynomial as $x^3-ax^2+bx-2010$, just written in a form that shows the roots directly.
💡 A polynomial's roots are exactly the numbers that make each factor zero, so writing it as a product of $(x-\text{root})$ pieces puts the roots in plain sight.
7.EE.A.1 Step 2 Match coefficients to pin down a
- Multiply out $(x-p)(x-q)(x-r)$ and line up the pieces with $x^3-ax^2+bx-2010$.
- The constant term is $-pqr$, and the $x^2$ term is $-(p+q+r)x^2$.
- Matching them term by term shows that the three roots multiply to $2010$ and that $a$ equals their sum.
- So minimizing $a$ means minimizing the sum of three positive integers whose product is fixed at $2010$.
💡 Expanding the factored form and comparing like terms turns the hidden coefficient $a$ into something concrete: the sum of the roots.
4.OA.B.4 Step 3 Break 2010 into its prime factors
- To build three whole-number roots that multiply to $2010$, first break $2010$ into primes.
- Divide step by step: $2010=2\cdot1005=2\cdot3\cdot335=2\cdot3\cdot5\cdot67$.
- That is four prime factors, but we need only three roots, so two of these primes must be multiplied together to form a single root.
💡 Every whole number is a fixed bag of primes, so any set of factors of $2010$ is just a way of dividing that bag $\{2,3,5,67\}$ into groups.
4.OA.B.4 Step 4 List every way to make three factors
- To get three roots we pick two of the four primes to merge into one root and leave the other two primes as the remaining two roots.
- There are only six choices for which pair to merge; for each, add up the three resulting factors:
💡 Listing all the groupings guarantees we do not miss a smaller sum hiding among the possibilities.
6.EE.B.5 Step 5 Pick the smallest sum
- The sum is smallest when the merged pair is the two smallest primes, $2$ and $3$, giving the root $6$ and leaving roots $5$ and $67$.
- That keeps the big prime $67$ from being pushed even higher, and it makes the three factors as balanced as the primes allow.
- The sum is $6+5+67=78$, so the smallest possible $a$ is $78$, which is choice (A).
💡 With a fixed product, a sum of factors shrinks when the factors are kept as balanced as possible, so merging the two smallest primes wastes the least.
6.EE.A.2 Call the three positive integer roots $p$, $q$, and $r$. A cubic whose leading c 7.EE.A.1 Multiply out $(x-p)(x-q)(x-r)$ and line up the pieces with $x^3-ax^2+bx-2010$. T 4.OA.B.4 To build three whole-number roots that multiply to $2010$, first break $2010$ in 4.OA.B.4 To get three roots we pick two of the four primes to merge into one root and lea 6.EE.B.5 The sum is smallest when the merged pair is the two smallest primes, $2$ and $3$ Review
Reasonableness: The three roots $6$, $5$, and $67$ multiply to $6\cdot5\cdot67=2010$, matching the constant term, and they are all positive integers, so this cubic really exists. Every other grouping gave a larger sum ($80$, $84$, and up), so $78$ is genuinely the minimum. It is also the smallest answer choice, exactly what we expect when the question asks for the smallest $a$. A tempting trap is pairing $30$ and $67$ (giving $1+30+67$ or the balanced-looking $30\cdot67=2010$ with sum $98$), but that ignores that merging the two smallest primes beats it.
Alternative: Instead of listing groupings, argue directly with the Extreme Principle: the prime $67$ must sit inside exactly one root, and any factor containing $67$ is already at least $67$. To keep the total down, leave $67$ alone as its own root and split the remaining product $2\cdot3\cdot5=30$ into the other two roots as evenly as possible, namely $6$ and $5$. That gives $67+6+5=78$ again.
CCSS standards used (min grade 7)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the three roots $p,q,r$ and writing the cubic as $(x-p)(x-q)(x-r)$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding the product of the three factors and matching coefficients to get $pqr=2010$ and $a=p+q+r$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Prime-factorizing $2010=2\cdot3\cdot5\cdot67$ and listing the ways to group the primes into three factors.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Comparing the candidate sums to find the value of $a$ that is smallest.)
⭐ The three roots multiply to $2010$ and $a$ is their sum, so split $2010$'s primes $2\cdot3\cdot5\cdot67$ into three factors as balanced as possible: $6$, $5$, $67$ give $a=78$.
⭐ The three roots multiply to $2010$ and $a$ is their sum, so split $2010$'s primes $2\cdot3\cdot5\cdot67$ into three factors as balanced as possible: $6$, $5$, $67$ give $a=78$.
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