AMC 10 · 2021 · #20

Grade 11 algebra
Archetype Modular Cycle Frequency Count └ Sub-archetype Power-Cycle Residue
polynomial-remainderdifference-of-cubesroots-of-unitymodular-arithmetic easier-related-problempattern-recognitionconvert-to-algebra ↑ Prerequisites: polynomial-remainderdifference-of-cubes
📏 Long solution 💡 3 insights
Problem
A very high-degree polynomial is divided by one quadratic. The remainder has degree less than two. Identify that remainder.

Pick an answer.

(A)
${-}z$
(B)
${-}1$
(C)
2021
(D)
z+1
(E)
2z+1
How to solve
Strategy Solve an Easier Related Problem

Nobody divides a degree-2021 polynomial by hand, so Tool #9 (Solve an Easier Related Problem) runs the show: find a much smaller polynomial that leaves the same remainder as z²⁰²¹+1, then divide that one instead. Tool #15 (Organize Information in More Ways) supplies the key re-arrangement — multiplying the divisor by z-1 turns z²+z+1 into the far friendlier z³-1. Tool #5 (Look for a Pattern) then reads off what that means: powers of z repeat in a 3-cycle once we work modulo the divisor, so only 2021 modulo 3 matters. Tool #4 (Introduce a Variable) frames the target as R(z)=az+b from the start, which keeps the degree condition front and center and tells us exactly when we are allowed to stop.

1STEP 1

Name the remainder's shape

The remainder is linear.

z²⁰²¹+1=(z²+z+1)Q(z)+R(z), R(z)=az+b
2STEP 2

Make the divisor useful

It divides the cube minus one.

(z-1)(z²+z+1)=z³-1 ⟹ (z²+z+1) ∣ (z³-1)
3STEP 3

Shrink the exponent

Trade the huge exponent for a small one.

z²⁰²¹-z²=z²((z³)⁶⁷³-1)=z²(z³-1)(z³·672+z³·671+…+z³+1)
4STEP 4

The difference is a multiple

The difference is a multiple of the divisor.

z²⁰²¹+1=(z²+1)+(z²⁰²¹-z²)_multiple of z²+z+1
5STEP 5

Finish the small division

The remainder is negative z.

z²+1=(z²+z+1) · 1+(-z) ⟹ R(z)=-z → (A)
Answer
-z
Test the claim with actual numbers by setting z=3. Then the divisor is 3²+3+1=13, and since 3³=27 leaves remainder 1 upon division by 13, we get 3²⁰²¹=(3³)⁶⁷³ · 3², which leaves remainder 9; adding 1 makes the dividend leave remainder 10 upon division by 13. Now evaluate each choice at z=3 and reduce modulo 13: (A) -3 leaves 10, (B) -1 leaves 12, (C) 2021 leaves 6, (D) 4 leaves 4, (E) 7 leaves 7. Only (A) matches, so the numeric check both confirms -z and eliminates all four rivals. A degree check backs this up: the remainder must be linear or constant, and -z is linear.
💡Key takeaway

When the divisor secretly says z³=1, a monstrous power collapses instantly: keep only what the exponent leaves over after dividing by 3.

  • Name what the remainder can be
  • Make the divisor useful
  • Shrink the exponent 2021 to 2
  • Trade the huge power for a small one
  • Finish the small division

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