The number of three-digit numbers divisible by 3 by considering the possible sums of digits that are divisible by 3. Your approach is as follows:

1. Sum of digits is 3: $(1, 1, 1)$ - 1 possible number

2. Sum of digits is 9: $(1, 3, 5)$ and $(3, 3, 3)$ -

Let's consider the cases separately :

a. Sum of digits is 9: $(1, 3, 5)$ For this case, we can arrange the digits in $3!$ ways : 135, 153, 315, 351, 513, and 531.

b. Sum of digits is 9: $(3, 3, 3)$ For this case, since all the digits are the same, there is only 1 possible number : 333.

Now, the total number of possible numbers when the sum of digits is 9 is :

$$ 3! + 1 = 6 + 1 = 7 $$

3. Sum of digits is 12: $(1, 3, 8)$ - $3!$ possible numbers

4. Sum of digits is 15: $(5, 5, 5)$ - 1 possible number

5. Sum of digits is 18: $(5, 5, 8)$ - $\frac{3!}{2!}$ possible numbers (since 5 is repeated)

6. Sum of digits is 21: $(5, 8, 8)$ - $\frac{3!}{2!}$ possible numbers (since 8 is repeated)

7. Sum of digits is 24: $(8, 8, 8)$ - 1 possible number

Adding up the possible numbers for each case, we get:

$$ 1 + 7 + 3! + 1 + \frac{3!}{2!} + \frac{3!}{2!} + 1 = 1 + 7 + 6 + 1 + 3 + 3 + 1 = 22 $$

So, there are a total of 22 three-digit numbers divisible by 3 that can be formed using the digits $1, 3, 5, 8$ with repetition allowed.