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.
• Permutation: The number of ways to arrange $n$ distinct objects is $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• Counting Principle: If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.

Step-by-Step Solution

Step 1: Identify Valid Sets of 5 Digits

We need to select 5 digits from the set $\{0, 1, 2, 3, 4, 5\}$ such that their sum is divisible by 3. The sum of all the digits in the set is $0 + 1 + 2 + 3 + 4 + 5 = 15$. Since we are forming a 5-digit number, we will exclude one digit, say $x$. For the resulting 5-digit number to be divisible by 3, $15 - x$ must be divisible by 3. Since 15 is divisible by 3, $x$ must also be divisible by 3. Therefore, we need to find which digits in the set $\{0, 1, 2, 3, 4, 5\}$ are divisible by 3.

The digits divisible by 3 are 0 and 3. This gives us two cases:

• Case 1: Exclude 0. The set of digits is $S_1 = \{1, 2, 3, 4, 5\}$. The sum of the digits is $1 + 2 + 3 + 4 + 5 = 15$, which is divisible by 3.

• Case 2: Exclude 3. The set of digits is $S_2 = \{0, 1, 2, 4, 5\}$. The sum of the digits is $0 + 1 + 2 + 4 + 5 = 12$, which is divisible by 3.

Step 2: Calculate the Number of 5-Digit Numbers for Each Valid Set

Now we calculate how many 5-digit numbers can be formed from each set.

• Case 1: Using the digits from $S_1 = \{1, 2, 3, 4, 5\}$

  Since none of the digits is 0, any permutation of these 5 digits will form a valid 5-digit number. The number of such permutations is $5!$.

  $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

  Therefore, 120 numbers can be formed using the digits $\{1, 2, 3, 4, 5\}$.

• Case 2: Using the digits from $S_2 = \{0, 1, 2, 4, 5\}$

  Here, we have the digit 0. This means the first digit cannot be 0. To find the number of valid 5-digit numbers, we first find the total number of permutations, and then subtract the number of permutations where 0 is the first digit.

  1. Total permutations of 5 digits: Ignoring the restriction on the first digit, there are $5!$ ways to arrange the 5 digits.

     $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

  2. Permutations starting with 0: If 0 is the first digit, there are 4 remaining digits $\{1, 2, 4, 5\}$ to arrange in the remaining 4 positions. This can be done in $4!$ ways.

     $4! = 4 \times 3 \times 2 \times 1 = 24$

  3. Valid 5-digit numbers: Subtract the number of permutations starting with 0 from the total number of permutations.

     $120 - 24 = 96$

  Therefore, 96 numbers can be formed using the digits $\{0, 1, 2, 4, 5\}$.

Step 3: Calculate the Total Number of Such Numbers

To find the total number of 5-digit numbers divisible by 3, we add the numbers from Case 1 and Case 2.

Total numbers = (Numbers from $S_1$) + (Numbers from $S_2$)

Total numbers = $120 + 96 = 216$

Common Mistakes & Tips

• Forgetting the Divisibility Rule: Always start by applying the divisibility rule to identify the possible sets of digits.
• Handling Zero Correctly: Remember that a number cannot start with zero. Subtract the cases where zero is in the first position.
• Systematic Approach: Be systematic in listing the possible cases to avoid missing any combinations.

Summary

We used the divisibility rule for 3 to find two sets of 5 digits from the given set whose sum is divisible by 3. Then, for each set, we calculated the number of 5-digit numbers that could be formed, making sure to handle the case where 0 was in the set correctly. Finally, we added the results from the two cases to get the total number of such numbers. The final answer is 216.

Final Answer

The final answer is $\boxed{216}$, which corresponds to option (D).