1991 AMC 12 第 26 题

先试着解答 1991 AMC 12 第 26 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 1991 AMC 12 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。

26.

若一个 nn 位正整数的 nn 个数字是集合 {1,2,…,n}\{1,2,\ldots,n\} 的一种排列,并且其前 kk 位组成的整数能被 kk 整除,其中 k=1k=1、22、…\ldots、nn,则称它为可爱数。例如,321321 是一个 33 位可爱数,因为 11 整除 33、22 整除 3232,且 33 整除 321321。共有多少个 66 位可爱数?
An nn-digit positive integer is cute if its nn digits are an arrangement of the set {1,2,…,n}\{1,2,\ldots,n\} and its first kk digits form an integer that is divisible by k,k, for k=1,k=1, 2,2, …,\ldots, n.n. For example, 321321 is a cute 33-digit integer because 11 divides 3,3, 22 divides 32,32, and 33 divides 321.321. How many cute 66-digit integers are there?
00
11
22
33
44
答案:C
知识点:整除性排列case analysis
小提示:
第二位必须为偶数,前三位的数字和必须能被 33 整除,前四位必须能被 44 整除
The second digit must be even, the third-prefix digit sum must be divisible by 3,3, and the fourth prefix must be divisible by 44
大提示:
第五位必须为 55,整个数必须为偶数且能被 33 整除
The fifth digit must be 5,5, and the full number must be even and divisible by 33
解答:
能被 55 整除迫使第五位为 55。能被 22 和 66 整除迫使第二位和第六位为偶数。接着依次应用整除规则:前三位的数字和必须能被 33 整除,由第三位和第四位组成的两位数必须能被 44 整除。检查集合 {1,2,3,4,6}\{1,2,3,4,6\} 中余下数字的排列,只剩 123654123654 和 321654321654。直接检验可知,这两个数都满足全部六个前缀整除条件,所以共有 22 个。
因此正确答案为 C。
Divisibility by 55 forces the fifth digit to be 5.5. Divisibility by 22 and 66 forces the second and sixth digits to be even. Now apply the divisibility tests successively: the first three digits must have sum divisible by 3,3, and the two-digit number formed by the third and fourth digits must be divisible by 4.4. Checking the remaining choices from {1,2,3,4,6}\{1,2,3,4,6\} leaves 123654123654 and 321654.321654. Each number directly satisfies all six prefix divisibility conditions, so there are 2.2.
Thus the correct answer is C.
LIVE by Po-Shen Loh
LIVE项目

LIVE直播课程

Self-Paced录播课程

启明星Star计划

奖学金项目

关于我们

罗博深是谁?

邮件联系

订阅英文简报

加入我们

申请明星导师

For a more Thought Full world.

© 2026 Expii, Inc.