AMC 10 · 2010 · #25
Grade 7 algebraLet a>0, and let P(x) be a polynomial with integer coefficients such that
P(1)=P(3)=P(5)=P(7)=a,
and
P(2)=P(4)=P(6)=P(8)=−a.
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: A polynomial $P(x)$ with integer coefficients satisfies $P(1)=P(3)=P(5)=P(7)=a$ and $P(2)=P(4)=P(6)=P(8)=-a$, where $a>0$. Find the smallest possible value of $a$.
Givens: $P(x)$ has integer coefficients; $P(1)=P(3)=P(5)=P(7)=a$; $P(2)=P(4)=P(6)=P(8)=-a$; $a>0$; Answer choices: (A) $105$, (B) $315$, (C) $945$, (D) $7!$, (E) $8!$
Unknowns: The smallest positive integer $a$ for which such a polynomial $P$ exists
Understand
Restated: A polynomial $P(x)$ with integer coefficients satisfies $P(1)=P(3)=P(5)=P(7)=a$ and $P(2)=P(4)=P(6)=P(8)=-a$, where $a>0$. Find the smallest possible value of $a$.
Givens: $P(x)$ has integer coefficients; $P(1)=P(3)=P(5)=P(7)=a$; $P(2)=P(4)=P(6)=P(8)=-a$; $a>0$; Answer choices: (A) $105$, (B) $315$, (C) $945$, (D) $7!$, (E) $8!$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #14 Extreme Principle, #3 Eliminate Possibilities
Tool #4 (Introduce a Variable): don't chase $P$ directly. Define the auxiliary polynomial $Q(x)=P(x)-a$, whose four odd roots let you factor and introduce a second unknown polynomial $R(x)$ with integer coefficients. Tool #7 (Subproblems): the four even inputs $2,4,6,8$ each give a separate divisibility condition on $2a$ — evaluate them one at a time. Tool #14 (Extreme Principle): the smallest $a$ is pinned down by the least common multiple of the four denominators. Tool #3 (Eliminate Possibilities): the choices are far apart, so a single divisibility failure kills $105$ and shows $315$ is minimal.
Execute — Answer: B
6.EE.B.5 Step 1 Shift to make roots
- Define a new polynomial $Q(x)=P(x)-a$.
- Because $P(1)=P(3)=P(5)=P(7)=a$, we get $Q(1)=Q(3)=Q(5)=Q(7)=0$, so $1,3,5,7$ are roots of $Q$.
- Since $P$ has integer coefficients and $a$ is an integer, $Q$ has integer coefficients too.
💡 Subtracting the common value $a$ turns the four odd inputs into genuine roots you can factor out.
6.EE.A.3 Step 2 Factor out the four roots
- The four roots give four factors.
- Since $(x-1)(x-3)(x-5)(x-7)$ is monic with integer coefficients, dividing $Q$ by it leaves an integer-coefficient quotient $R(x)$ with no remainder.
- Hence $Q(x)=(x-1)(x-3)(x-5)(x-7)\,R(x)$, so $P(x)=a+(x-1)(x-3)(x-5)(x-7)\,R(x)$ for some polynomial $R$ with integer coefficients.
💡 Each integer root $r$ lets you pull a factor $(x-r)$ out while keeping integer coefficients.
6.EE.A.2 Step 3 Use the even inputs
- Plug in the even inputs $x=2,4,6,8$, where $P(x)=-a$.
- Subtracting $a$ from both sides, the factored form must equal $-2a$ at each of them: $(x-1)(x-3)(x-5)(x-7)\,R(x)=-2a$.
💡 At each even input the known factor product must absorb $-2a$, leaving an integer value of $R$.
7.NS.A.2 Step 4 Evaluate the four products
- Compute the product of signed distances at each even input: at $x=2$, $(1)(-1)(-3)(-5)=-15$; at $x=4$, $(3)(1)(-1)(-3)=9$; at $x=6$, $(5)(3)(1)(-1)=-15$; at $x=8$, $(7)(5)(3)(1)=105$.
- So $-15\,R(2)=-2a$, $9\,R(4)=-2a$, $-15\,R(6)=-2a$, and $105\,R(8)=-2a$.
💡 Multiplying the four signed distances gives the exact number that must divide $2a$ at that input.
4.OA.B.4 Step 5 Turn into divisibility
- Each $R$-value must be an integer, so every denominator must divide $2a$: we need $15\mid 2a$, $9\mid 2a$, and $105\mid 2a$ (the $x=6$ condition repeats the $x=2$ one).
- These are the only constraints, because once they hold the required values $R(2),R(4),R(6),R(8)$ are integers that a valid integer polynomial can hit.
💡 An integer output means each denominator has to be a factor of $2a$.
6.NS.B.4 Step 6 Minimize with the LCM
- For all three to hold, $2a$ must be a multiple of $\operatorname{lcm}(9,15,105)$.
- Since $9=3^2$, $15=3\cdot5$, and $105=3\cdot5\cdot7$, the least common multiple is $3^2\cdot5\cdot7=315$.
- So $315\mid 2a$.
- Because $315$ is odd, it shares no factor of $2$, hence $315\mid a$.
- The smallest positive such $a$ is $315$.
💡 The least $a$ is forced by the least common multiple of all the denominators.
4.OA.B.4 Step 7 Confirm and match the choice
- Check $a=315$ actually works: then $2a=630$, giving $R(2)=42$, $R(4)=-70$, $R(6)=42$, $R(8)=-6$, all integers, and an integer-coefficient polynomial $R$ through these four points exists, so a valid $P$ exists.
- Thus $315$ is both necessary and achievable, and the smallest possible value of $a$ is $315$, which is choice (B).
💡 $a=315$ meets every divisibility rule and is reachable, so nothing smaller can work.
6.EE.B.5 Define a new polynomial $Q(x)=P(x)-a$. Because $P(1)=P(3)=P(5)=P(7)=a$, we get $ 6.EE.A.3 The four roots give four factors. Since $(x-1)(x-3)(x-5)(x-7)$ is monic with int 6.EE.A.2 Plug in the even inputs $x=2,4,6,8$, where $P(x)=-a$. Subtracting $a$ from both 7.NS.A.2 Compute the product of signed distances at each even input: at $x=2$, $(1)(-1)(- 4.OA.B.4 Each $R$-value must be an integer, so every denominator must divide $2a$: we nee 6.NS.B.4 For all three to hold, $2a$ must be a multiple of $\operatorname{lcm}(9,15,105)$ 4.OA.B.4 Check $a=315$ actually works: then $2a=630$, giving $R(2)=42$, $R(4)=-70$, $R(6) Review
Reasonableness: Test the nearby choices against the divisibility rules. Choice (A) $105$: then $2a=210$, and $210/9$ is not an integer, so the $x=4$ condition $9\mid 2a$ fails — $105$ is missing the second factor of $3$. Choice (C) $945$: $2a=1890$ is divisible by $9$, $15$, and $105$, so it works, but $945=3\times315$ is three times too big. Choices (D) $7!=5040$ and (E) $8!=40320$ are far larger still. So $315$ is the least value that clears every condition, matching the explicit construction $R(2)=42,\ R(4)=-70,\ R(6)=42,\ R(8)=-6$.
Alternative: Tool #5 (Look for a Pattern) via finite differences. The eight values sit at equally spaced inputs $1,\ldots,8$ and alternate $a,-a,a,-a,\ldots$. Building the table of successive differences of this data, integer coefficients require each order-$k$ difference to be divisible by the corresponding factorial-type factor; the binding condition again reduces to $2a$ being a multiple of $315$. Same answer, reached by differencing the data instead of factoring.
CCSS standards used (min grade 7)
6.EE.B.5Understand solving an equation or inequality as a process of finding values (Recognizing that $x=1,3,5,7$ make $Q(x)=P(x)-a$ equal to zero, i.e. are roots.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring $Q(x)$ as $(x-1)(x-3)(x-5)(x-7)R(x)$ and rewriting $P(x)$ in factored form.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating the factored form of $P$ at the even inputs $x=2,4,6,8$.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Computing the signed products $(1)(-1)(-3)(-5)=-15$, etc., and the resulting quotients for $R$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Turning integer-value requirements into the divisibility conditions $15\mid 2a$, $9\mid 2a$, $105\mid 2a$ and eliminating $105$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Combining the conditions via $\operatorname{lcm}(9,15,105)=315$ to find the smallest $a$.)
⭐ Even a scary final problem cracks open once you subtract $a$ to create roots, factor out $(x-1)(x-3)(x-5)(x-7)$, and let integer coefficients force $2a$ to be a multiple of $\operatorname{lcm}(9,15,105)=315$ — just grade-7 factors and multiples.
⭐ Even a scary final problem cracks open once you subtract $a$ to create roots, factor out $(x-1)(x-3)(x-5)(x-7)$, and let integer coefficients force $2a$ to be a multiple of $\operatorname{lcm}(9,15,105)=315$ — just grade-7 factors and multiples.
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