# 2011 AMC 8 Problems/Problem 23

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? $\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$

## 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 $48+36=\boxed{\textbf{(D)}\ 84}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 