2021 AMC 12A Spring Problem 8

Attempt Problem 8 of the 2021 AMC 12A Spring below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Spring solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

A sequence of numbers is defined by D0=0,D_0 = 0, D1=0,D_1 = 0, D2=1,D_2 = 1, and Dn=Dn−1+Dn−3D_n = D_{n-1} + D_{n-3} for n≥3.n \ge 3. What are the parities (evenness or oddness) of the triple of numbers (D2021,D2022,D2023),(D_{2021}, D_{2022}, D_{2023}), where EE denotes even and OO denotes odd?
(O,E,O)(O, E, O)
(E,E,O)(E, E, O)
(E,O,E)(E, O, E)
(O,O,E)(O, O, E)
(O,O,O)(O, O, O)
Answer: C
Concepts:parityrecursionpattern recognition
Small hint:
Only parities matter, so compute the sequence with each term as EE or OO using O+O=EO+O=E
Big hint:
The parity sequence repeats with period 7;7; reduce 2021,2022,20232021, 2022, 2023 modulo 77
Solution:
Working modulo 2,2, the terms D0,D1,D2,…D_0, D_1, D_2, \ldots have parities E,E, E,E, O,O, O,O, O,O, E,E, O,O, E,E, E,E, O,O, O,O, O,O, E,E, O,…O, \ldots which repeat with period 77 starting from D0D_0 (indeed D7,D8,D9D_7, D_8, D_9 have the same parities E,E,OE, E, O as D0,D1,D2D_0, D_1, D_2).
Since 2021≡5,2021 \equiv 5, 2022≡6,2022 \equiv 6, and 2023≡0(mod7),2023 \equiv 0 \pmod 7, the parities match those of D5,D6,D0,D_5, D_6, D_0, namely E,O,E.E, O, E.
Thus, the correct answer is C.
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