2011
Problem - 1172
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?
When the units digit is $5$, the thousands digits has $4$ choices (cannot be zero), the hundreds and tens digits have $4$ and $3$ choices respectively (including $0$). Therefore, the number of such integers is $$1\times 4\times 4\times 3={48}$$
When the units digit is $0$, the digit $5$ has three places to choose from. Then the remaining two places have $4$ and $3$ choices of digits, respectively. Therefore, the number of such integers is $$1\times 3\times 4\times 3 = 36$$
Therefore, the final answer is $$48+36=\boxed{84}$$