2023 AMC 8 第 18 题

先试着解答 2023 AMC 8 第 18 题,然后核对你的答案与视频与精心整理的解答,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 8 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。

18.

蚱蜢 Greta 坐在池塘中一长排睡莲叶上。从任意一片睡莲叶出发,Greta 可以向右跳 55 片,或向左跳 33 片。Greta 至少要跳多少次,才能到达从起点向右 20232023 片的睡莲叶?

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump 55 pads to the right or 33 pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located 20232023 pads to the right of her starting position?

405405

407407

409409

411411

413413

答案:D
知识点:丢番图方程模运算最优化
难度评级:1740
视频讲解:
解答视频缩略图
Play video

Click to load, then click again to play

文字解答:

设向右跳 rr 次,向左跳 ll 次,则 模 553l3(mod5)-3l\equiv3\pmod5,所以 l4(mod5)l\equiv4\pmod55r3l=2023. 5r-3l=2023.

为了最少跳跃,取满足条件的最小 ll,也就是 l=4l=4。此时 5r12=20235r-12=2023,所以 r=407r=407

总跳数为 407+4=411407+4=411

所以正确答案是 D

Let rr be the number of right jumps and ll be the number of left jumps. We need 5r3l=2023. 5r-3l=2023. Modulo 55, this gives 3l3(mod5)-3l\equiv3\pmod5, so l4(mod5)l\equiv4\pmod5.

To minimize the total number of jumps, use the smallest possible ll, namely l=4l=4. Then 5r12=20235r-12=2023, so r=407r=407.

The fewest number of jumps is 407+4=411407+4=411.

Thus, D is the correct answer.

← 第 17 题#17
完整试卷

其他年份的第 18 题

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8