Key Concepts and Formulas

• Divisibility Rule for 3: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
• Counting Principle: If there are $n_1$ ways to do one thing and $n_2$ ways to do another, then there are $n_1 \times n_2$ ways to do both.

Step-by-Step Solution

Step 1: Classify the Digits Based on their Remainder when Divided by 3

We are given the digits $1, 3, 5, 8$. We want to classify them based on their remainders when divided by 3.

• $1 \equiv 1 \pmod{3}$ (Remainder is 1)
• $3 \equiv 0 \pmod{3}$ (Remainder is 0)
• $5 \equiv 2 \pmod{3}$ (Remainder is 2)
• $8 \equiv 2 \pmod{3}$ (Remainder is 2)

So, we have the following classification:

• Remainder 0: $\{3\}$ (1 digit)
• Remainder 1: $\{1\}$ (1 digit)
• Remainder 2: $\{5, 8\}$ (2 digits)

Step 2: Determine the Possible Combinations of Remainders for a Sum Divisible by 3

Let the three digits be $d_1, d_2, d_3$. For the three-digit number to be divisible by 3, the sum of the digits $d_1 + d_2 + d_3$ must be divisible by 3. We consider the possible combinations of remainders (modulo 3) that sum to a multiple of 3. The possible combinations are:

• Case 1: All three digits have a remainder of 0 (0+0+0).
• Case 2: All three digits have a remainder of 1 (1+1+1).
• Case 3: All three digits have a remainder of 2 (2+2+2).
• Case 4: One digit has a remainder of 0, one has a remainder of 1, and one has a remainder of 2 (0+1+2).

Step 3: Calculate the Number of Three-Digit Numbers for Each Case

• Case 1: All three digits have a remainder of 0. The only digit with a remainder of 0 is 3. So, the number must be 333. There is only 1 such number.

• Case 2: All three digits have a remainder of 1. The only digit with a remainder of 1 is 1. So, the number must be 111. There is only 1 such number.

• Case 3: All three digits have a remainder of 2. The digits with a remainder of 2 are 5 and 8. So, each of the three digits can be either 5 or 8. There are $2 \times 2 \times 2 = 8$ such numbers.

• Case 4: One digit has a remainder of 0, one has a remainder of 1, and one has a remainder of 2. The digits are 3, 1, and either 5 or 8.

  • If the digits are 3, 1, and 5, there are $3! = 6$ permutations.
  • If the digits are 3, 1, and 8, there are $3! = 6$ permutations. So, there are $6 + 6 = 12$ such numbers.

Step 4: Calculate the Total Number of Three-Digit Numbers

The total number of three-digit numbers divisible by 3 is $1 + 1 + 8 + 12 = 22$.

Common Mistakes & Tips

• Remember the divisibility rule for 3.
• Don't forget to consider all possible combinations of remainders that sum to a multiple of 3.
• Be careful when calculating permutations versus combinations. In this case, the order of the digits matters, so we use permutations.

Summary

We classified the given digits based on their remainders when divided by 3. Then, we identified all possible combinations of remainders that result in a sum divisible by 3. For each combination, we calculated the number of three-digit numbers that can be formed using the given digits. Finally, we summed the counts from each case to find the total number of three-digit numbers divisible by 3. The total number of such numbers is 22.

Final Answer

The final answer is \boxed{22}, which corresponds to option (B).