To find how many 3-digit positive integers are odd and do not contain the digit 5, we can break down the problem step by step.
A 3-digit positive integer has the format XYZ, where:
- X represents the hundreds place
- Y represents the tens place
- Z represents the units (or ones) place
1. **Identifying the units digit (Z):** Since we want odd integers, Z can only be one of the digits from the set {1, 3, 7, 9}. There are 4 options here (1, 3, 7, and 9) because 5 is not allowed.
2. **Identifying the hundreds digit (X):** The hundreds digit can be any digit from 1 to 9 except 5. Thus, the possible digits for X are {1, 2, 3, 4, 6, 7, 8, 9}, which gives us 8 choices in total.
3. **Identifying the tens digit (Y):** The tens digit can be any digit from 0 to 9 except 5. Therefore, the possible choices for Y are {0, 1, 2, 3, 4, 6, 7, 8, 9}, totaling 9 options.
Next, we can calculate the total number of valid combinations:
Number of possibilities = (Choices for X) × (Choices for Y) × (Choices for Z) = 8 (for X) × 9 (for Y) × 4 (for Z) = 288.
So, the total number of 3-digit positive integers that are odd and do not contain the digit 5 is 288.