AMC8 2011

AMC8 2011 · Q23

AMC8 2011 · Q23. It mainly tests Basic counting (rules of product/sum), Casework.

How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

有多少个四位正整数具有四个不同数字，开头数字不为零，该整数是5的倍数，且5是其中最大的数字？

(A) 24 24

(B) 48 48

(D) 84 84

(E) 108 108

Answer

Correct choice: (D)

正确答案：（D）

Solution

We can separate this into two cases. If an integer is a multiple of $5,$ the last digit must be either $0$ or $5.$ Case 1: The last digit is $5.$ The leading digit can be $1,2,3,$ or $4.$ Because the second digit can be $0$ but not the leading digit, there are also $4$ choices. The third digit cannot be the leading digit or the second digit, so there are $3$ choices. The number of integers is this case is $4\cdot4\cdot3\cdot1=48.$ Case 2: The last digit is $0.$ Because $5$ is the largest digit, one of the remaining three digits must be $5.$ There are $3$ ways to choose which digit should be $5.$ The remaining digits can be $1,2,3,$ or $4,$ but since they have to be different there are $4\cdot3$ ways to choose. The number of integers in this case is $1\cdot3\cdot4\cdot3=36.$ Therefore, the answer is $48+36=\boxed{\textbf{(D)}\ 84}$.

可以分为两种情况。如果一个整数是$5$的倍数，最后一位必须是$0$或$5$。情况1：最后一位是$5$。首位可以是$1,2,3$或$4$，有$4$种选择。第二位可以是$0$，也有$4$种选择。第三位不能是首位或第二位，有$3$种选择。此情况数量是$4\cdot4\cdot3\cdot1=48$。情况2：最后一位是$0$。因为$5$是最大数字，剩下三位中必须有一个是$5$，有$3$种选择位置。剩下两位从$1,2,3,4$中选不同数字，有$4\cdot3$种。此情况数量是$1\cdot3\cdot4\cdot3=36$。总计$48+36=\boxed{\textbf{(D)}\ 84}$。

Topics