Question

The number of natural numbers less than 10000 which are divisible by 5 and that no digit is repeated in the same number, is

• A. 802
• B. 602
• C. 702
• D. 1169

Hint:

When solving such problems, break it down by case (for 2-digit, 3-digit, and 4-digit numbers) and consider the divisibility rules and digit constraints.

Answer 1

The Correct Option is D

We need to find how many numbers less than 10000 are divisible by 5 and have no repeated digits. The number must be divisible by 5, so the last digit must be either 0 or 5.
Case 1: Numbers of 4 digits (1000 to 9999)
- The last digit must be 0 or 5.
- If the last digit is 0, we can choose the first three digits from 9 digits (1 to 9), the second digit from 8 digits, and the third digit from 7 digits, so there are \( 9 \times 8 \times 7 = 504 \) numbers.
- If the last digit is 5, we can choose the first three digits from 9 digits (1 to 9, excluding 5), the second digit from 8 digits, and the third digit from 7 digits, so there are \( 9 \times 8 \times 7 = 504 \) numbers.
Thus, there are \( 504 + 504 = 1008 \) numbers.
Case 2: Numbers of 3 digits (100 to 999)
- The last digit must be 0 or 5.
- If the last digit is 0, we can choose the first two digits from 9 digits (1 to 9), and the second digit from 8 digits, so there are \( 9 \times 8 = 72 \) numbers. - If the last digit is 5, we can choose the first two digits from 9 digits (1 to 9, excluding 5), and the second digit from 8 digits, so there are \( 9 \times 8 = 72 \) numbers.
Thus, there are \( 72 + 72 = 144 \) numbers.
Case 3: Numbers of 2 digits (10 to 99)
- The last digit must be 0 or 5.
- If the last digit is 0, we can choose the first digit from 9 digits (1 to 9), so there are 9 numbers.
- If the last digit is 5, we can choose the first digit from 8 digits (1 to 9, excluding 5), so there are 8 numbers.
Thus, there are \( 9 + 8 = 17 \) numbers.
Total: The total number of numbers is: \[ 1008 + 144 + 17 = 1169 \] Thus, the correct answer is option (4).