Q. 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?

Explanation : 

How many 3 digit natural numbers ??

Given conditions:
– Must be a 3-digit number
– All digits must be odd
– No repetition of digits allowed
– Number must be divisible by 5

Solution steps:

1. For divisibility by 5:
– Last digit must be either 0 or 5
– Since only odd digits allowed, last digit must be 5

2. Available odd digits: 1, 3, 5, 7, 9
– 5 is already used in units place
– Remaining digits for other positions: 1, 3, 7, 9

3. Filling other positions:
– Hundreds place: 4 choices (1, 3, 7, 9)
– Tens place: 3 choices (remaining digits after hundreds place)

4. Total possible numbers:
– Using multiplication principle
– 4 × 3 = 12 different numbers

Answer: 12 three-digit numbers possible