This is a problem from a "SIMAK UI 2010" exam, asking about how many possible employee numbers (nomor pegawai) that are odd. The employee number consists of three digits, and the first digit cannot be zero.

We need to count how many of these employee numbers are odd.

Key Points:

1. The employee number has three digits.
2. The first digit cannot be zero.
3. We are asked to count how many of these numbers are odd. A number is odd if its last digit is odd.

Step-by-Step Solution:

• Last digit (odd): The last digit must be odd. Possible odd digits are {1, 3, 5, 7, 9}, so there are 5 choices for the last digit.

• First digit (non-zero): The first digit must be non-zero and can be any of {1, 2, 3, 4, 5, 6, 7, 8, 9}, so there are 9 choices for the first digit.

• Middle digit: The middle digit can be any of the 10 digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, so there are 10 choices for the middle digit.

Total number of odd employee numbers:

The total number of possible three-digit employee numbers that are odd is:

$9 \text{ (choices for the first digit)} \times 10 \text{ (choices for the middle digit)} \times 5 \text{ (choices for the last digit)} = 450$

Thus, the correct answer is 450.

Answer: D. 450

Would you like a detailed breakdown of a similar problem? Here are five related questions to consider:

1. How many three-digit numbers exist where the first digit is not zero?
2. What is the total number of even three-digit numbers with the first digit non-zero?
3. How does the restriction on the first digit (not zero) affect the total possible outcomes?
4. How would the answer change if no restrictions were placed on the first digit?
5. How many employee numbers would be divisible by 5?

Tip: When solving problems involving digits and restrictions, it's helpful to break the problem into cases and consider how many choices you have at each step.