AMC 10 · 2010 · #21

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 whole-coefficient polynomial takes one value at four inputs and its negative at four others. Find the smallest such value.

Pick an answer.

(A)

105

(B)

315

(C)

945

(D)

(E)

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

Shifting turns four inputs into roots.

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

2STEP 2

Factor out the four roots

Those roots factor straight out.

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

3STEP 3

Use the even inputs

The other four inputs give four products.

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

4STEP 4

Evaluate the four products

Each product must stay a whole number.

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

5STEP 5

Turn into divisibility

So three divisibility demands appear.

15 ∣ 2a, 9 ∣ 2a, 105 ∣ 2a

6STEP 6

Minimize with the LCM

Their least common multiple gives 315.

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

7STEP 7

Confirm and match the choice

A real polynomial reaches it, so 315 stands.

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