How many four-digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

To solve this problem, we need to calculate how many four-digit numbers divisible by 6 can be formed using the digits \(0, 2, 3, 4, 6\), where no digit is repeated and \(0\) is not the first digit. 

A number is divisible by 6 if it satisfies two conditions:
1. **Divisibility by 2**: The last digit must be even.
2. **Divisibility by 3**: The sum of the digits must be divisible by 3.

Let's go step by step:

### Step 1: Available digits
The digits we can use are \(0, 2, 3, 4, 6\), and there are five available digits. We are tasked with creating a four-digit number, so we will select four of these digits.

### Step 2: Condition for divisibility by 2
Since the number must be divisible by 2, the last digit must be an even digit. The available even digits are \(0, 2, 4, 6\). We will select the last digit from this set.

### Step 3: Condition for divisibility by 3
For divisibility by 3, the sum of the digits of the number must be divisible by 3. The sum of all available digits (\(0, 2, 3, 4, 6\)) is:
\[
0 + 2 + 3 + 4 + 6 = 15
\]
Since 15 is divisible by 3, removing any digit will still result in a sum divisible by 3, meaning we don’t need to worry about this condition—it will automatically be satisfied.

### Step 4: Selecting the last digit
Now, let’s calculate the number of possible four-digit numbers based on each case for the last digit:

- **Case 1: Last digit is 0**  
  The first digit cannot be 0, and we have \(2, 3, 4, 6\) left to choose from for the first three digits.
  - The first digit can be chosen from \{2, 3, 4, 6\}, so there are 4 choices.
  - The second digit can be chosen from the remaining 3 digits.
  - The third digit can be chosen from the remaining 2 digits.
  So, the total number of possibilities in this case is:
  \[
  4 \times 3 \times 2 = 24
  \]

- **Case 2: Last digit is 2**  
  The remaining digits are \{0, 3, 4, 6\}.
  - The first digit cannot be 0, so it must be chosen from \{3, 4, 6\}, giving 3 choices.
  - The second digit can be chosen from the remaining 3 digits.
  - The third digit can be chosen from the remaining 2 digits.
  So, the total number of possibilities in this case is:
  \[
  3 \times 3 \times 2 = 18
  \]

- **Case 3: Last digit is 4**  
  The remaining digits are \{0, 2, 3, 6\}.
  - The first digit cannot be 0, so it must be chosen from \{2, 3, 6\}, giving 3 choices.
  - The second digit can be chosen from the remaining 3 digits.
  - The third digit can be chosen from the remaining 2 digits.
  So, the total number of possibilities in this case is:
  \[
  3 \times 3 \times 2 = 18
  \]

- **Case 4: Last digit is 6**  
  The remaining digits are \{0, 2, 3, 4\}.
  - The first digit cannot be 0, so it must be chosen from \{2, 3, 4\}, giving 3 choices.
  - The second digit can be chosen from the remaining 3 digits.
  - The third digit can be chosen from the remaining 2 digits.
  So, the total number of possibilities in this case is:
  \[
  3 \times 3 \times 2 = 18
  \]

### Step 5: Total possibilities
Summing the possibilities from all four cases:
\[
24 + 18 + 18 + 18 = 78
\]

Thus, the total number of four-digit numbers divisible by 6 that can be formed using the digits \(0, 2, 3, 4, 6\) without repetition and with no \(0\) in the left-most position is \( \boxed{78} \).

Would you like a detailed explanation for any specific part?

Here are five related questions:
1. How would the solution change if repetition of digits was allowed?
2. What if the number had to be divisible by 5 instead of 6?
3. How many three-digit numbers divisible by 6 can be formed using the same set of digits?
4. How would you handle the case where the first digit can be zero?
5. What if you were allowed to use five digits instead of four?

**Tip**: Always check divisibility rules carefully, as they simplify the process of eliminating invalid numbers.

Discover how to form four-digit numbers divisible by 6 using the digits 0, 2, 3, 4, and 6 without repetition. Learn how the divisibility rules for 2 and 3 apply, with a detailed step-by-step solution. Suitable for Grades 9-12.

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