Question

N>40000, where N is divisible by 5. How many such 5 digit numbers can be formed using 0,1,3,5,7,9 without repetition.

Answer 1

The correct answer is 120.

Answer 2

We are to form 5-digit numbers using the digits {0, 1, 3, 5, 7, 9} without repetition, under two conditions:

• The number must be greater than 40,000.
• The number is divisible by 5 (i.e. its last digit must be 0 or 5)

Case 1: Last Digit is 0

With the last digit fixed as 0, the first digit cannot be 0 and must make the number > 40,000. The available digits for the first position (excluding 0) are {1, 3, 5, 7, 9}. Since the number must be > 40,000, the first digit must be at least 4. Among our choices, only 5, 7, and 9 qualify.

Number of choices for the first digit: 3.

After fixing the first and last digits, there remain 4 digits for the three middle positions. The number of ways to fill these positions is:

P(4, 3) = 4 × 3 × 2 = 24

Total numbers for Case 1: 3 × 24 = 72.

Case 2: Last Digit is 5

With the last digit fixed as 5, the first digit must be chosen from the remaining digits {0, 1, 3, 7, 9} (0 is not allowed in the first position) and must be at least 4 to ensure the number is > 40,000. This leaves only 7 and 9.

Number of choices for the first digit: 2.

The remaining three positions (the 2nd, 3rd, and 4th digits) can be filled from the 4 remaining digits (from a total of 6, after fixing the first and last digits) in:

P(4, 3) = 4 × 3 × 2 = 24

Total numbers for Case 2: 2 × 24 = 48.

Total Count

Total 5-digit numbers = 72 + 48 = 120.