Question

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?

$24$

$48$

$60$

$84$

$108$

Solution:

For a number to be divisible by $5,$ the units digit must be either $0$ or $5.$

If the units digit is $0,$ one of the other three digits must be $5.$ The remaining two digits must be chosen from $\{1, 2, 3, 4\}.$ There are $6$ ways to choose the pair, and there are $6$ ways to arrange the three digits for a total of $6 \cdot 6 = 36$ numbers.

If the units digit is $5,$ there are $4$ ways to choose the thousands digit. There are $4 \cdot 3 = 12$ ways to choose the other $2$ digits. This leaves a total of $4 \cdot 12 = 48$ numbers for this case.

Combining both cases, we get the total number of such integers is $36 + 48 = 84.$

Thus, D is the correct answer.

2011 AMC 8 Problem 23

Solution:

Problem 23 in Other Years