To solve this problem, we need to determine how many five-digit odd numbers are possible, given that the leftmost digit cannot be zero.

Steps to Solve:

1. First Digit (Leftmost Digit):

   • Since the leftmost digit cannot be zero, it can be any digit from 1 to 9.
   • So, there are 9 possible choices for the first digit.

2. Last Digit (Odd Number Requirement):

   • The number must be odd, so the last digit must be 1, 3, 5, 7, or 9.
   • There are 5 possible choices for the last digit.

3. Middle Three Digits:

   • Each of these three digits (second, third, and fourth positions) can be any digit from 0 to 9.
   • So, there are 10 possible choices for each of these three digits.

Total Number of Five-Digit Odd Numbers:

• Multiply the number of choices for each digit: $\text{Total} = 9 \times 10 \times 10 \times 10 \times 5 = 45000$

Thus, there are 45,000 possible five-digit odd numbers where the leftmost digit cannot be zero.

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Would you like more details on this process, or do you have any questions?

Related Questions:

1. How many five-digit even numbers are possible if the leftmost digit cannot be zero?
2. How many six-digit odd numbers are possible if the leftmost digit cannot be zero?
3. How many five-digit numbers (without the odd requirement) are possible if the leftmost digit cannot be zero?
4. How many five-digit numbers are possible if there are no restrictions on any digit?
5. How many four-digit odd numbers are possible if the leftmost digit cannot be zero?

Tip: Always start by identifying the constraints for each digit in a multi-digit number problem.