What is the largest two-digit number that is divisible by both of its digits, and the digits are distinct?
Since we are looking for the largest number, we should start at 99 and work our way down.  No number in the 90s works, because the only numbers divisible by $9$ are $90$ and $99$. $90$ is invalid because no number is divisible by zero, and $99$ is invalid because the digits are the same.  The same reasoning applies to numbers with $8$, $7$, $6$, or $5$ in the tens place.  When we get to the 40s, however, there are three numbers that are divisible by $4$: $40$, $44$, and $48$. $48$ is also divisible by $8$, so the number we are looking for is $\boxed{48}.$