How many natural numbers less than 600 & divisible by 5can be formed using 0, 1, 2, 3, 4, 5, 9(a) Repetition of digits not allowed

Sure, let's solve this step by step:

Step 1: Identify the conditions We need to find natural numbers less than 600 that are divisible by 5 and can be formed using the digits 0, 1, 2, 3, 4, 5, 9. Repetition of digits is not allowed.

Step 2: Identify the possible number of digits The numbers can be 1-digit, 2-digit, or 3-digit numbers. 4-digit numbers are not possible as they would exceed 600.

Step 3: Calculate for 1-digit numbers The only 1-digit number divisible by 5 is 5 itself. So, there is 1 such number.

Step 4: Calculate for 2-digit numbers The 2-digit numbers divisible by 5 must end in 5 or 0. But since we can't start a number with 0, the only 2-digit numbers we can form are those that end in 5. The first digit can be 1, 2, 3, or 4. So, there are 4 such numbers.

Step 5: Calculate for 3-digit numbers The 3-digit numbers divisible by 5 must also end in 5 or 0. But since we can't start a number with 0, the only 3-digit numbers we can form are those that end in 5. The first digit can be 1, 2, 3, 4, or 5, and the second digit can be any of the remaining 5 digits. So, there are 5*5 = 25 such numbers.

Step 6: Add up the numbers Adding up the numbers from steps 3, 4, and 5, we get 1 + 4 + 25 = 30. So, there are 30 natural numbers less than 600 that are divisible by 5 and can be formed using the digits 0, 1, 2, 3, 4, 5, 9 without repetition.