Divisibility by $15$ .

For a number to be a multiple of 15, it has to be divisible by both 3 and 5. For 3, divisibility can be checked by adding the sum of digits which must be a multiple of 3. Since the number consists of only 8s and 0s, we need to find smallest multiple of 8 divisible by 3, i.e. 3*8. Thus the number needs to contain three 8s and one zero in the end to be a multiple of 5. Hence we obtain the number : 8880.

We know that the multiples of 15 are 15, 30, 45, 60...

Given that we can not have any 5s, we know that every other term is immediately removed. Because of that we know the number must be a multiple of 30 (which is the same as saying it must be a multiple of both 3 and 10)

That means it must end in a 0 and the sum of the digits must be divisible by 3 thus we have...

88888...80

we want the smallest number that does this, and the fewest number of 8s we can have to still meet the conditions set is 3

so we find 8880

80, 800, 808, 880, 8000, 8800, 8080 are not divisible by 15. 8880 is the first number to be divisible by 15 containing only 8s and 0

The number's got to end with a zero because it ending with 8 won't make it a multiple of 5, and hence, not 15. Secondly, the number's also got to be divisible by 3, and hence, we'll have to write as less 8's as possible to make it divisible by 3. So the number is 8880.

80/15 gives you a remainder of 5/15 or 1/3. Increasing to 880 gives a remainder of 2/3. The pattern continues giving you no remainder every third term. 80, 880, 8880 , 88880, 888880. 8888880