AMC 10 · 2010 · #25

Grade 7 algebranumber-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

polynomial-rootspolynomial-factoringlcmdivisibility-rules extreme-principle ↑ Prerequisites: polynomial-factoring

📏 Long solution 💡 4 insights

Problem

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 is greater than 0. Find the smallest possible value of a.

Pick an answer.

(A)

105

(B)

315

(C)

945

(D)

(E)

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

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.

1STEP 1

Shift to make roots

Set Q(x)=P(x)-a. Since P equals a at 1, 3, 5, 7, those four are roots of Q, and Q still has integer coefficients.

Q(x)=P(x)-a, Q(1)=Q(3)=Q(5)=Q(7)=0

2STEP 2

Factor out the four roots

The monic integer factor (x-1)(x-3)(x-5)(x-7) divides Q exactly, so P(x)=a+(x-1)(x-3)(x-5)(x-7)R(x) for an integer-coefficient R.

P(x)=a+(x-1)(x-3)(x-5)(x-7) R(x), RinZ[x]

3STEP 3

Use the even inputs

At x=2, 4, 6, 8 we have P(x)=-a, so the factored part must equal -2a at each of those four inputs.

(x-1)(x-3)(x-5)(x-7) R(x)=-2a at x=2,4,6,8

4STEP 4

Evaluate the four products

The four signed products are -15, 9, -15 and 105, so -15R(2)=9R(4)=-15R(6)=105R(8)=-2a.

R(2)=2a/15, R(4)=-2a/9, R(6)=2a/15, R(8)=-2a/105

5STEP 5

Turn into divisibility

Each R value must be an integer, so 2a needs 15, 9 and 105 as divisors — the x=6 condition just repeats x=2.

15 ∣ 2a, 9 ∣ 2a, 105 ∣ 2a

6STEP 6

Minimize with the LCM

lcm(9, 15, 105)=3²·5·7=315 must divide 2a, and since 315 is odd it must already divide a itself.

lcm(9,15,105)=3²·5·7=315, 315 ∣ 2a→ 315 ∣ a

7STEP 7

Confirm and match the choice

At a=315 the values R(2)=42, R(4)=-70, R(6)=42, R(8)=-6 are all integers, so a valid P exists — choice (B).

a=315: R(2)=42, R(4)=-70, R(6)=42, R(8)=-6 ✓

Answer

315

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 ∣ 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=3×315 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.

💡Key takeaway

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 lcm(9,15,105)=315 — just grade-7 factors and multiples.

Shift to make roots
Factor out the four roots
Use the even inputs
Evaluate the four products
Turn into divisibility
Minimize with the LCM
Confirm and match the choice