4-Digit Numbers Divisible by 4 Rule Application

The question asks us to find the count of 4-digit numbers that can be formed using the digits 1, 2, 3, and 4 exactly once (no repetition) such that the number is divisible by 4.

Divisibility Rule for 4

A key rule in number theory is that a number is divisible by 4 if the number formed by its last two digits (the tens and units place) is divisible by 4. We will use this rule.

Possible Last Two Digits for Divisibility by 4

We need to find pairs of distinct digits from the set {1, 2, 3, 4} that form a 2-digit number divisible by 4. Let's list the possibilities:

• Possible endings using 1: 12 (Yes, divisible by 4), 13 (No), 14 (No)
• Possible endings using 2: 21 (No), 23 (No), 24 (Yes, divisible by 4)
• Possible endings using 3: 31 (No), 32 (Yes, divisible by 4), 34 (No)
• Possible endings using 4: 41 (No), 42 (No), 43 (No)

So, the only possible pairs for the last two digits (units and tens place) are 12, 24, and 32.

Number Formation Based on Last Digits

Now, for each valid pair of last two digits, we need to determine how many ways the first two digits can be arranged using the remaining digits. Since repetition is not allowed, the first two digits must be the remaining two digits from the set {1, 2, 3, 4}.

Case 1: Last two digits are 12

The digits 1 and 2 are used for the tens and units places. The remaining digits are 3 and 4. These two digits (3 and 4) can be arranged in the first two positions (thousands and hundreds) in \(2!\) ways.

• The possible arrangements for the first two digits are 34 and 43.
• This gives us the numbers: 3412 and 4312.
• The number of ways for this case is \(2! = 2 \times 1 = 2\).

Case 2: Last two digits are 24

The digits 2 and 4 are used. The remaining digits are 1 and 3. These can be arranged in the first two positions in \(2!\) ways.

• The possible arrangements for the first two digits are 13 and 31.
• This gives us the numbers: 1324 and 3124.
• The number of ways for this case is \(2! = 2 \times 1 = 2\).

Case 3: Last two digits are 32

The digits 3 and 2 are used. The remaining digits are 1 and 4. These can be arranged in the first two positions in \(2!\) ways.

• The possible arrangements for the first two digits are 14 and 41.
• This gives us the numbers: 1432 and 4132.
• The number of ways for this case is \(2! = 2 \times 1 = 2\).

Total Count Calculation for Divisible Numbers

To find the total number of 4-digit numbers that meet the criteria, we add up the number of ways from each case:

Total Numbers = (Ways ending in 12) + (Ways ending in 24) + (Ways ending in 32)

Total Numbers = \(2 + 2 + 2 = 6\).

Final Count of Numbers

Therefore, there are exactly 6 distinct 4-digit numbers that can be formed using the digits 1, 2, 3, and 4 without repetition, and which are divisible by 4.