AMC 10 · 2022 · #11

Grade 11 algebra
complex-numbersroots-of-unityperiodic-functionexponentsmodular-arithmetic pattern-recognitioneasier-related-problemsystematic-enumeration ↑ Prerequisites: complex-numbersexponents
📏 Medium solution 💡 2 insights
Problem
Two fixed complex numbers are given. A rule raises each to the n-th power and adds the two results. Find the value at 2022, without ever expanding that power.

Pick an answer.

(A)
-2
(B)
-1
(C)
0
(D)
$\sqrt{3}$
(E)
2
How to solve
Strategy Look for a Pattern

Nobody expands a 2022nd power. So the real question is what these two bases do when you keep multiplying them by themselves, and tool #5 (Look for a Pattern) is the whole engine: if the powers repeat on a short cycle, then f repeats on that same cycle and a huge exponent costs nothing. Tool #9 (Solve an Easier Related Problem) is how the pattern gets found — replace n=2022 with n=1,2,3, which is arithmetic a person can actually do, and see what turns up. Tool #4 (Introduce a Variable) keeps the writing honest: give the two bases the names ω and ω so that facts about them can be stated once and reused, instead of dragging (-1+i√(3))/2 through every line. Tool #15 (Organize Information in More Ways) does the last lift: once the cycle length is known, 2022 has to be re-read not as a size but as a multiple of that cycle length, and the answer falls out of the rewriting.

1STEP 1

Name the two bases

The two are conjugates.

ω = -1/2 + √(3)/2i, ω = -1/2 - √(3)/2i, f(n) = ωⁿ + ωⁿ
2STEP 2

Square it

Squaring gives the other one.

ω² = ((-1+i√(3))²)/4 = (1 - 2i√(3) - 3)/4 = (-2-2i√(3))/4 = (-1-i√(3))/2 = ω, ω² = ω
3STEP 3

Cube it

Cubing gives one.

ω³ = ω · ω² = ω · ω = (1+3)/4 = 1, ω³ = 1
4STEP 4

The values repeat every three

They repeat every three.

ωⁿ⁺³ = ωⁿ · ω³ = ωⁿ ⟹ f(n+3) = f(n), and for n = 0,1,2,3,4,5,6,7,8 the values are f(n) = 2, -1, -1, 2, -1, -1, 2, -1, -1
5STEP 5

Reduce 2022 modulo three

The digit sum shows it is a multiple of three.

2+0+2+2 = 6 = 3 × 2 ⟹ 3 ∣ 2022, 2022 = 3 × 674, 2022 ≡ 0 (mod 3)
6STEP 6

Collapse the power

Both become one, so the sum is 2.

f(2022) = (ω³)⁶⁷⁴ + (ω³)⁶⁷⁴ = 1⁶⁷⁴ + 1⁶⁷⁴ = 1 + 1 = 2 (E)
Answer
2
Two checks, one structural and one numerical. Structurally, ω sits on the unit circle: |ω| = √((-1/2)² + (√(3)/2)²) = √(1/4+3/4) = 1, so every power of ω also has modulus 1 and f(n) = 2 Re(ωⁿ) can never leave the range from -2 to 2. Since f(2022) came out at the very top of that range, it says ω²⁰²² is exactly 1 — the strongest possible statement, and consistent with 2022 being a multiple of 3. Numerically, ω ≈ -0.5 + 0.8660i; computing ωⁿ + ωⁿ in decimals for n = 0 through 8 returns 2, -1, -1, 2, -1, -1, 2, -1, -1 with zero imaginary part every time, and evaluating the 2022nd powers directly returns 2.000. The wrong choices are all traceable. Choice (B) -1 is the answer for any n that is not a multiple of 3, so it catches anyone who mis-divides 2022. Choice (C) 0 assumes the two conjugate terms cancel, which is what happens to their imaginary parts but never to their real parts. Choice (D) √(3) is a fragment of the given numbers, not a value f ever takes. Choice (A) -2 would require ωⁿ = -1, impossible because ω has odd order 3. Only (E) 2 survives.
💡Key takeaway

When a base returns to 1 after a few multiplications, its powers run in a short loop — so a monstrous exponent like 2022 only ever asks one question: what is the remainder?

  • Name the two bases
  • Square omega and read the result
  • Cube it and land on 1
  • The values repeat every three
  • Read 2022 as a multiple of 3
  • Collapse the power and finish