Forming Four-Digit Numbers Divisible by 4
This problem involves finding the number of unique four-digit numbers that can be created using the digits \(1, 2, 3, 5\) without repetition. The key condition is that these numbers must be divisible by \(4\). Let's break this down step-by-step.
Understanding the Divisibility Rule for 4
A fundamental rule in mathematics states that an integer is divisible by \(4\) if the number formed by its last two digits (the tens and units digits) is itself divisible by \(4\).
The set of digits we can use is \(\{1, 2, 3, 5\}\). We need to find pairs from this set that can form the last two digits of our four-digit number.
Identifying Valid Two-Digit Endings
We need to find two-digit numbers formed using digits from \(\{1, 2, 3, 5\}\) without repetition, which are divisible by \(4\). Let's list the possibilities:
| Tens Digit | Units Digit | Two-Digit Number | Divisible by 4? |
|---|---|---|---|
| 1 | 2 | 12 | Yes (\(12 = 4 \times 3\)) |
| 1 | 3 | 13 | No |
| 1 | 5 | 15 | No |
| 2 | 1 | 21 | No |
| 2 | 3 | 23 | No |
| 2 | 5 | 25 | No |
| 3 | 1 | 31 | No |
| 3 | 2 | 32 | Yes (\(32 = 4 \times 8\)) |
| 3 | 5 | 35 | No |
| 5 | 1 | 51 | No |
| 5 | 2 | 52 | Yes (\(52 = 4 \times 13\)) |
| 5 | 3 | 53 | No |
The possible two-digit endings that are divisible by \(4\) are \(12, 32,\) and \(52\).
Calculating Permutations for Each Valid Ending
We have identified three possible endings (\(12, 32, 52\)). For each ending, we need to determine how many ways the first two digits can be arranged using the remaining unused digits.
Case 1: The number ends with 12.
Digits used: \(1, 2\). Remaining digits: \(\{3, 5\}\).
The first two positions of the four-digit number must be filled by arranging \(3\) and \(5\). The number of permutations is \(2!\) (2 factorial).
Calculation: \(2! = 2 \times 1 = 2\). The numbers formed are \(3512\) and \(5312\).
Case 2: The number ends with 32.
Digits used: \(3, 2\). Remaining digits: \(\{1, 5\}\).
The first two positions must be filled by arranging \(1\) and \(5\). The number of permutations is \(2!\).
Calculation: \(2! = 2 \times 1 = 2\). The numbers formed are \(1532\) and \(5132\).
Case 3: The number ends with 52.
Digits used: \(5, 2\). Remaining digits: \(\{1, 3\}\).
The first two positions must be filled by arranging \(1\) and \(3\). The number of permutations is \(2!\).
Calculation: \(2! = 2 \times 1 = 2\). The numbers formed are \(1352\) and \(3152\).
Total Count of Numbers Divisible by 4
To find the total count, we sum the number of possibilities from each case:
Total Count = (Count for ending 12) + (Count for ending 32) + (Count for ending 52)
Total Count = \(2 + 2 + 2 = 6\).
Thus, there are \(6\) four-digit numbers formed using the digits \(1, 2, 3, 5\) without repetition that are divisible by \(4\).
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