Problem Analysis

We need to find the number of 5-digit numbers that can be formed using the digits {1, 2, 3, 4, 5, 6} without repetition, such that the number is divisible by 4.

1. Applying the Divisibility Rule for 4

A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Our first step is to identify all possible pairs of last two digits from the given set {1, 2, 3, 4, 5, 6} that form a number divisible by 4.

The possible endings are:

• 12
• 16
• 24
• 32
• 36
• 52
• 56
• 64

There are a total of 8 possible pairs for the last two digits.

2. Arranging the Remaining Digits

We are forming a 5-digit number. Once we fix the last two digits, we need to fill the first three positions.

• For each of the 8 endings, we have used 2 out of the 6 available digits.
• This leaves us with $6 - 2 = 4$ digits to fill the first three places.
• The number of ways to arrange 3 digits from a set of 4 available digits is a permutation, calculated as $P(4, 3)$.

$P(4, 3) = \frac{4!}{(4-3)!} = 4! = 4 \times 3 \times 2 = 24$ ways.

This means for each of the 8 possible endings, there are 24 different ways to arrange the first three digits.

3. Calculating the Total Number of Possibilities

The total number of 5-digit numbers (n) is the product of the number of possible endings and the number of ways to arrange the remaining digits.

n = (Number of possible endings) × (Number of ways to fill the first three places)

n = 8 × 24 = 192

Conclusion

The value of n is 192.