How many four-digit numbers which are divisible by 12 can be formed using the digits 0, 2, 4, 5 and 6 such that no digit is used more than once and 0 is not used in the leftmost position?

Sure, let's solve this step by step:

Step 1: Understand the problem We need to find out how many four-digit numbers can be formed using the digits 0, 2, 4, 5, and 6, such that each digit is used only once, the number is divisible by 12, and 0 is not the first digit.

Step 2: Identify the conditions for a number to be divisible by 12 A number is divisible by 12 if it is divisible by both 3 and 4. For a number to be divisible by 3, the sum of its digits must be divisible by 3. For a number to be divisible by 4, the number formed by the last two digits must be divisible by 4.

Step 3: Find the possible combinations The sum of the digits 0, 2, 4, 5, and 6 is 17, which is not divisible by 3. Therefore, we must exclude one digit to make the sum divisible by 3. The only way to do this is to exclude the 5, as 17 - 5 = 12, which is divisible by 3.

Now, we have the digits 0, 2, 4, and 6 left. We need to form a four-digit number that is divisible by 4. This means the last two digits must be 04, 20, 40, or 60.

Step 4: Count the combinations If the last two digits are 04, the first digit can be 2 or 6 (2 options), and the second digit can be the remaining one (1 option). So, there are 2 * 1 = 2 combinations.

If the last two digits are 20, 40, or 60, the first digit can be 4 or 6 (2 options), and the second digit can be the remaining one (1 option). So, there are 3 * 2 * 1 = 6 combinations.

Step 5: Add up the combinations The total number of four-digit numbers that can be formed is 2 + 6 = 8.