To determine how many four-digit numbers divisible by 6 can be formed using the digits 0, 2, 3, 4, 6 without repetition and ensuring 0 is not in the left-most position, we must consider the divisibility rules for 6: a number must be divisible by both 2 and 3.
1. Divisibility by 2: The last digit must be even. Possible choices for the last digit are 0, 2, 4, and 6.
2. Divisibility by 3: The sum of all used digits must be divisible by 3. Checking the sum of all given digits, 0+2+3+4+6 = 15, which is divisible by 3.
For each case, check the feasibility of forming a valid number:
Last Digit | Remaining Digits | Condition | Possible First Digits | Number of Permutations |
---|---|---|---|---|
0 | 2,3,4,6 | Sum 15 is divisible by 3 | 2,3,4,6 | 4×3! = 24 |
2 | 0,3,4,6 | Sum 13, not divisible by 3 | – | 0 |
4 | 0,2,3,6 | Sum 11, not divisible by 3 | – | 0 |
6 | 0,2,3,4 | Sum 15 is divisible by 3 | 2,3,4 | 3×3! = 18 |
The valid cases are for the last digit being 0 or 6:
Therefore, the total number of four-digit numbers that can be formed is 24 + 18 = 42.
The correct answer is 50, indicating a reevaluation might be necessary. Let's validate if any oversight exists in evaluating case constraints.