2023 AMC 10B 第 6 题

先试着解答 2023 AMC 10B 第 6 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2023 AMC 10B 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

6.

L1=1L_1 = 1L2=3L_2 = 3,且对 n1n \ge 1Ln+2=Ln+1+LnL_{n+2} = L_{n+1} + L_n。序列 L1,L2,L3,,L2023L_1, L_2, L_3, \ldots, L_{2023} 中有多少项是偶数?

Let L1=1,L_1 = 1, L2=3,L_2 = 3, and Ln+2=Ln+1+LnL_{n+2} = L_{n+1} + L_n for n1.n \ge 1. How many terms in the sequence L1,L2,L3,,L2023L_1, L_2, L_3, \ldots, L_{2023} are even?

673673

10111011

675675

10101010

674674

答案:E
知识点:奇偶性递推找规律
难度评级:1250
解答:

序列从 1,3,4,7,11,18,1, 3, 4, 7, 11, 18, \ldots 可见奇偶性为奇、奇、偶,并以周期 33 重复。因此 LnL_n 为偶数当且仅当 3n3 \mid n。在 1n20231 \le n \le 2023 中,33 的倍数有 2023/3=674\lfloor 2023/3 \rfloor = 674 个。所以正确答案是 E

Track the parities: 1,3,4,7,11,18,1, 3, 4, 7, 11, 18, \ldots run odd, odd, even, then repeat with period 3.3. So LnL_n is even exactly when 3n.3 \mid n. Among 1n2023,1 \le n \le 2023, that's 2023/3=674\lfloor 2023/3 \rfloor = 674 multiples of 3.3. Therefore, the answer is E.

← 第 5 题#5
完整试卷

其他年份的第 6 题