AMC 10 · 2021 · #8

Grade 4 pattern
Archetype Modular Cycle Frequency Count └ Sub-archetype Position-in-Cycle Lookup
recursive-sequenceparitymodular-arithmeticpattern-recognition pattern-recognitionsystematic-enumeration ↑ Prerequisites: parity
📏 Medium solution 💡 2 insights
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Problem
A sequence starts with zero, zero, one, and from then on every term is the previous term plus the term three places back. Say whether the 2021st, 2022nd, and 2023rd terms are each even or odd, in that order.

Pick an answer.

(A)
(O,E,O)
(B)
(E,E,O)
(C)
(E,O,E)
(D)
(O,O,E)
(E)
(O,O,O)
How to solve
Strategy Look for a Pattern

Tool #9 (Easier Related Problem) makes the whole thing tractable: throw away the actual sizes and keep only even/odd, because parity of a sum is decided by parity of the parts. Tool #2 (Systematic List) builds the even/odd table from n = 0 upward without skipping a term. Tool #5 (Look for a Pattern) is the payoff — the parity list repeats, and once the pattern's length is known, index 2021 collapses to a tiny index. Tool #3 (Eliminate Possibilities) confirms the resulting triple against the five listed options.

1STEP 1

Track parity only

Track parity, not size.

E+E=E, E+O=O, O+O=E
2STEP 2

Build the first terms

Build the first dozen terms.

D₀,…,D₁₀ = 0, 0, 1, 1, 1, 2, 3, 4, 6, 9, 13
3STEP 3

Write the parity row

Rewrite each term as even or odd.

n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 ; D_n & 0 & 0 & 1 & 1 & 1 & 2 & 3 & 4 & 6 & 9 & 13 ; parity & E & E & O & O & O & E & O & E & E & O & O
4STEP 4

Spot the period

The parity repeats every seven.

(D₇,D₈,D₉) ≡ (E,E,O) ≡ (D₀,D₁,D₂) → parity of D_n = parity of D_n+7
5STEP 5

Reduce the indices

Reduce the three indices modulo seven.

2021 = 7 · 288 + 5, 2022 = 7 · 288 + 6, 2023 = 7 · 289 + 0
6STEP 6

Read the answer off the table

The answer is even, odd, even.

D₂₀₂₁ ≡ D₅ = E, D₂₀₂₂ ≡ D₆ = O, D₂₀₂₃ ≡ D₀ = E → (C) (E,O,E)
Answer
(E,O,E)
The period-7 claim is checkable beyond the one match used to prove it: parities at n = 10,11,12,13 are O,O,E,O (from 13, 19, 28, 41), exactly matching n = 3,4,5,6, which were O,O,E,O. Extending the table to n = 20 keeps the block E,E,O,O,O,E,O repeating with no drift. The three answer indices land on three different positions of the cycle, and the resulting triple (E,O,E) appears exactly once among the five options, so no other choice is compatible. As a sanity note on plausibility, D₂₀₂₃ = D₂₀₂₂ + D₂₀₂₀, and an even result requires its two addends to share a parity — consistent with the cycle, which puts D₂₀₂₂ at position 6 (O) and D₂₀₂₀ at position 4 (O).
💡Key takeaway

This AMC 12 problem needs no algebra beyond Grade 4: keep only even/odd instead of the huge numbers, list the first few terms, notice the pattern repeats every 7 steps, then divide 2021 by 7 and look up the remainder.

  • Track parity, not size
  • Build the first terms
  • Write the parity row
  • Spot the repeat: period 7
  • Reduce 2021, 2022, 2023 mod 7
  • Read the answer off the table

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