Question

The number of integers greater than 5000 and divisible by 5 that can be formed with the digits 1, 3, 5, 7, 8, 9 where no digit is repeated is

• A. 240
• B. 180
• C. 120
• D. 276

Hint:

In permutation problems with multiple constraints, always handle the most restrictive positions first. Here, the units digit (divisibility) and the first digit (range) are the most constrained, so they should be filled first.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a counting problem using principles of permutations. We need to count the number of possible integers that can be formed under several constraints: range (greater than 5000), divisibility, available digits, and no repetition.
Step 2: Key Formula or Approach:
We should break down the problem into cases based on the number of digits the integer can have. The constraints on the first and last digits must be handled carefully due to the overlapping use of digits.
Step 3: Detailed Explanation:
The available digits are \{1, 3, 5, 7, 8, 9\}. Constraint 1: Divisible by 5 The last digit must be 5. So, the units place is fixed. Available digits for other places: \{1, 3, 7, 8, 9\}. Constraint 2: Greater than 5000 This means the number can have 4, 5, or 6 digits. Case 1: 4-digit numbers The number has the form _ _ _ 5.

Units place: Fixed as 5 (1 choice).
Thousands place: Must be greater than or equal to 5 to make the number>5000. The available digits are \{1, 3, 7, 8, 9\}. So, the choices are \{7, 8, 9\} (3 choices). (Note: 5 is already used).
Hundreds place: We have used two digits (5 and one from \{7,8,9\}). From the original 6 digits, 4 are left. (4 choices).
Tens place: Now 3 digits are left. (3 choices).
Number of 4-digit numbers = \( 3 \times 4 \times 3 \times 1 = 36 \). Case 2: 5-digit numbers The number has the form _ _ _ _ 5. Any 5-digit number formed from these digits will be greater than 5000.

Units place: Fixed as 5 (1 choice).
Remaining 4 places: We need to arrange the remaining 5 digits \{1, 3, 7, 8, 9\} in 4 places.
Number of ways = \( P(5, 4) = \frac{5!}{(5-4)!} = 5! = 120 \). Case 3: 6-digit numbers The number has the form _ _ _ _ _ 5. Any 6-digit number formed from these digits will be greater than 5000.

Units place: Fixed as 5 (1 choice).
Remaining 5 places: We need to arrange the remaining 5 digits \{1, 3, 7, 8, 9\} in 5 places.
Number of ways = \( P(5, 5) = 5! = 120 \). Total Number of Integers Total = (Number of 4-digit numbers) + (Number of 5-digit numbers) + (Number of 6-digit numbers) Total = 36 + 120 + 120 = 276.
Step 4: Final Answer:
The total number of such integers is 276.