The [integer](https://artofproblemsolving.com/wiki/index.php/Integer) $n$ is the smallest [positive](https://artofproblemsolving.com/wiki/index.php/Positive) [multiple](https://artofproblemsolving.com/wiki/index.php/Multiple) of $15$ such that every [digit](https://artofproblemsolving.com/wiki/index.php/Digit) of $n$ is either $8$ or $0$. Compute $\frac{n}{15}$.
Any multiple of 15 is a multiple of 5 and a multiple of 3. 
Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units [digit](https://artofproblemsolving.com/wiki/index.php/Digit) of $n$ must be 0.  


The sum of the digits of any multiple of 3 must be [divisible](https://artofproblemsolving.com/wiki/index.php/Divisible) by 3.  If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$.  For this number to be divisible by 3, $a$ must be divisible by 3. We also know that $a>0$ since $n$ is positive. Thus $n$ must have at least three copies of the digit 8.  
The smallest number which meets these two requirements is 8880.  Thus the answer is $\frac{8880}{15} = \boxed{592}$.