How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?

Here's a brief explanation: 1. **Identify Odd Digits:** The odd digits are {1, 3, 5, 7, 9}. 2. **Divisibility by 5:** For a number to be divisible by 5, its last digit must be 0 or 5. Since all digits must be odd, the last digit (units place) must be 5. So, the number looks like _ _ 5. (1 choice for the last digit: 5) 3. **No Repetition & Remaining Digits:** We have used the digit 5. The remaining odd digits available for the first two places are {1, 3, 7, 9}. There are 4 such digits. 4. **First Digit (Hundreds Place):** We can choose any of the 4 remaining odd digits for the hundreds place. (4 choices) 5. **Second Digit (Tens Place):** After choosing the first digit, we have 3 odd digits left for the tens place (since repetition is not allowed). (3 choices) Total number of such 3-digit numbers = (Choices for Hundreds) x (Choices for Tens) x (Choices for Units) = 4 x 3 x 1 = 12 Therefore, there are 12 such numbers. The final answer is B) 12.