The total number of three-digit numbers, divisible by 3, which can be formed using the digits $1,3,5,8$, if repetition of digits is allowed, is :

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: <br/><br/>1. Sum of digits is 3: $(1, 1, 1)$ - 1 possible number <br/><br/>2. Sum of digits is 9: $(1, 3, 5)$ and $(3, 3, 3)$ - <br/><br/> Let's consider the cases separately : <br/><br/><b>a.</b> 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. <br/><br/><b>b.</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. <br/><br/>Now, the total number of possible numbers when the sum of digits is 9 is : <br/><br/>$3! + 1 = 6 + 1 = 7$ <br/><br/>3. Sum of digits is 12: $(1, 3, 8)$ - $3!$ possible numbers <br/><br/>4. Sum of digits is 15: $(5, 5, 5)$ - 1 possible number <br/><br/>5. Sum of digits is 18: $(5, 5, 8)$ - $\frac{3!}{2!}$ possible numbers (since 5 is repeated) <br/><br/>6. Sum of digits is 21: $(5, 8, 8)$ - $\frac{3!}{2!}$ possible numbers (since 8 is repeated) <br/><br/>7. Sum of digits is 24: $(8, 8, 8)$ - 1 possible number <br/><br/>Adding up the possible numbers for each case, we get: <br/><br/>$$ 1 + 7 + 3! + 1 + \frac{3!}{2!} + \frac{3!}{2!} + 1 = 1 + 7 + 6 + 1 + 3 + 3 + 1 = 22 $$ <br/><br/>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.