2021 AMC 12A Spring 第 8 题

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8.

数列由 D0=0D_0 = 0D1=0D_1 = 0D2=1D_2 = 1,以及对 n3n \ge 3 成立的 Dn=Dn1+Dn3D_n = D_{n-1} + D_{n-3} 定义。三元组 (D2021,D2022,D2023)(D_{2021}, D_{2022}, D_{2023}) 中各数的奇偶性是什么?其中 EE 表示偶数,OO 表示奇数。

A sequence of numbers is defined by D0=0,D_0 = 0, D1=0,D_1 = 0, D2=1,D_2 = 1, and Dn=Dn1+Dn3D_n = D_{n-1} + D_{n-3} for n3.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)

答案:C
知识点:奇偶性递推找规律
难度评级:1600
解答:

22 计算,项 D0,D1,D2,D_0, D_1, D_2, \ldots 的奇偶性为 EEEEOOOOOOEEOOEEEEOOOOOOEEO,O, \ldots,从 D0D_0 开始以 77 为周期重复,因为 D7,D8,D9D_7, D_8, D_9 的奇偶性与 D0,D1,D2D_0, D_1, D_2 相同,都是 E,E,OE, E, O

由于 202152021 \equiv 5202262022 \equiv 6,且 20230(mod7)2023 \equiv 0 \pmod 7,所求奇偶性分别与 D5,D6,D0D_5, D_6, D_0 相同,即 E,O,EE, O, E

因此,正确答案是 C

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 20215,2021 \equiv 5, 20226,2022 \equiv 6, and 20230(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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