To solve this problem, we need to determine how many 4-digit numbers can be created using the digits $0, 1, 2, \ldots, 9$ under the following conditions:

1. The number must be odd.
2. The number must be greater than 6000.
3. Repetition of digits is allowed.

Step 1: Determine the constraints for the digits.

• First digit (thousands place): The number must be greater than 6000, so the first digit can be $6, 7, 8,$ or $9$. This gives us 4 possible choices.
• Last digit (units place): The number must be odd, so the last digit must be $1, 3, 5, 7,$ or $9$. This gives us 5 possible choices.
• Middle two digits (hundreds and tens places): There are no restrictions on the hundreds and tens digits, so they can be any of the 10 digits $0, 1, 2, \ldots, 9$. Each digit has 10 possible choices.

Step 2: Calculate the total number of 4-digit numbers.

Now, we can calculate the total number of such 4-digit numbers by multiplying the number of choices for each digit:

$\text{Total number of 4-digit numbers} = (\text{choices for 1st digit}) \times (\text{choices for 2nd digit}) \times (\text{choices for 3rd digit}) \times (\text{choices for 4th digit})$

$\text{Total} = 4 \times 10 \times 10 \times 5 = 2000$

Final Answer:

There are 2000 different 4-digit numbers that can be created under the given conditions.

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Would you like further details or have any other questions? Here are some related problems to consider:

1. How many 4-digit even numbers greater than 6000 can be formed with repetition allowed?
2. If repetition is not allowed, how many 4-digit odd numbers greater than 6000 can be formed?
3. What if the number must be divisible by 5 and greater than 8000?
4. How many 4-digit numbers can be formed if the number must be less than 6000 and even?
5. What if the first digit cannot be 7 or 8, and the number must still be odd and greater than 6000?

Tip: When calculating the number of possible combinations, breaking the problem into smaller steps based on constraints makes it easier to manage complex conditions.