To determine how many three-digit positive integers are odd and do not contain the digit 5, we can break this down step by step.
A three-digit integer can be represented in the form ABC, where:
- A is the hundreds place digit (1-9),
- B is the tens place digit (0-9),
- C is the units (or ones) place digit, which must be odd.
Step 1: Identify allowable digits
The digits we can use (without including the digit 5) are: 0, 1, 2, 3, 4, 6, 7, 8, 9.
Step 2: Determine the options for each digit
- A (hundreds place): Since A must be from 1 to 9 but cannot be 5, the possible options are 1, 2, 3, 4, 6, 7, 8, 9. This gives us a total of 8 choices.
- B (tens place): B can be any digit from 0 to 9, excluding 5. This allows us to select from 0, 1, 2, 3, 4, 6, 7, 8, 9, giving us 9 possible choices.
- C (units place): To ensure the integer is odd, C can only be 1, 3, 7, or 9. This gives us 4 options for the units place digit.
Step 3: Calculate the total number of combinations
To find the total number of three-digit positive integers that are odd and without the digit 5, we multiply the number of choices for each digit:
Total = (Choices for A) × (Choices for B) × (Choices for C)
Total = 8 × 9 × 4 = 288
Conclusion
Thus, the total number of three-digit positive integers that are odd and do not contain the digit 5 is 288.