Question

How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition such that the number is divisible by 5?

• A. 24
• B. 48
• C. 60
• D. 120

Hint:

For divisibility by 5, the last digit must be 0 or 5. When digits don't include 0, last digit must be 5. Arrange remaining digits without repetition.

Answer 1

The Correct Option is A

Digits available are \(\{1, 2, 3, 4, 5\}\). The 4-digit number must be divisible by 5. A number is divisible by 5 if its last digit is 0 or 5. Since 0 is not available, the last digit must be 5. Step 1: Fix the last digit as 5. Step 2: The first three digits are to be chosen from \(\{1, 2, 3, 4\}\) without repetition. Step 3: Number of ways to arrange the first three digits = \[ P(4,3) = 4 \times 3 \times 2 = 24 \] Step 4: Total number of such 4-digit numbers = \(24 \times 1 = 24\).

Answer 2

To solve the problem, we need to find how many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition such that the number is divisible by 5. 

1. Understanding Divisibility by 5:
A number is divisible by 5 if its last digit is either 0 or 5.
Since the digits given are 1, 2, 3, 4, 5 and 0 is not included, the last digit must be 5 for the number to be divisible by 5.

2. Fixing the Last Digit:
Last digit = 5 (fixed)

3. Selecting the First Three Digits:
The remaining digits available for the first three positions are {1, 2, 3, 4} since digits cannot be repeated.

4. Counting the Number of Possible Arrangements:
Number of ways to arrange the first digit = 4 (any of the four digits)
Number of ways to arrange the second digit = 3 (remaining digits)
Number of ways to arrange the third digit = 2 (remaining digits)

5. Calculating Total Number of 4-digit Numbers:
Total numbers = $4 \times 3 \times 2 = 24$

Final Answer:
The number of 4-digit numbers divisible by 5 that can be formed without repetition is $ {24} $.