AMC 10 · 2022 · #20

Grade 8 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

polynomial-factoringpolynomial-rootssystems-of-equationspolynomial-remainderconvert-to-algebra easier-related-problemconvert-to-algebraidentify-subproblems ↑ Prerequisites: polynomial-factoringsystems-of-equations

📏 Long solution 💡 4 insights

Problem

A polynomial with rational coefficients leaves a prescribed linear remainder when divided by each of two quadratics. Among all such polynomials, exactly one has the smallest degree. Find the sum of the squares of its coefficients.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem) — first try the smallest possible degree for P (degree 2, with constant quotients) to see what goes wrong; then bump up to degree 3 (linear quotients). Tool #7 (Subproblems) — splitting equation A = B into three coefficient-equations (one per power of x) turns the polynomial puzzle into a small system. Tool #13 (Algebra) — solve that tiny system for the unknown coefficients. Tool #3 (Eliminate) — quickly verify the final sum 1²+2²+3²+3² = 23 matches choice (E).

1STEP 1

Rule out degree two

Degree two is impossible.

degree 2 → 3 = 2 (impossible)

2STEP 2

Let each quotient be linear

Both quotients are linear.

Q₁ = ax+b, Q₂ = ax+c

3STEP 3

Expand the first expression

Expand the first expression.

P = ax³ + (a+b)x² + (a+b+1)x + (b+2)

4STEP 4

Expand the second expression

Expand the second too.

P = ax³ + cx² + (a+2)x + (c+1)

5STEP 5

Match the coefficients

Matching coefficients fixes the unknowns.

a = 1, b = 1, c = 2

6STEP 6

Write the polynomial

The polynomial is complete.

P(x) = x³ + 2x² + 3x + 3

7STEP 7

Sum the squared coefficients

Squaring and adding gives 23.

1² + 2² + 3² + 3² = 23 → (E)

Answer

Sanity. Verify P(x) = x³ + 2x² + 3x + 3 in both original conditions. Divide by x² + x + 1: P = (x²+x+1)(x+1) + (x+2), remainder x+2 — matches. Divide by x² + 1: P = (x²+1)(x+2) + (2x+1), remainder 2x+1 — matches. Sum of squared coefficients 1+4+9+9 = 23, which is exactly choice (E). The degree-2 attempt failed by a numerical contradiction (3 = 2), so degree 3 is genuinely the least. The choices 10, 13, 19, 20, 23 are decoys for forgetting a coefficient or for the unsquared sum 1+2+3+3 = 9.

💡Key takeaway

This AMC 12 problem only needs Grade 8 systems of equations you already know — try the smallest possible degree first (it fails by a clean 3=2 contradiction), then write the next degree using letters for the unknown coefficients and match like terms to set up three tiny equations. Solve to get P(x) = x³ + 2x² + 3x + 3, and the sum of squares of coefficients is 1+4+9+9 = 23.