Question

Using digits 1 to 6 (each at most once), how many 4-digit numbers can be formed that are divisible by 4?

Hint:

For permutation problems with divisibility constraints, always start by filling the places that are restricted by the rule (e.g., the last digit for divisibility by 2 or 5, the last two for divisibility by 4).

Answer 1

Correct Answer: 96

Step 1: Understanding the Question: 
We need to form 4-digit numbers using the digits \(\{1, 2, 3, 4, 5, 6\}\) without repetition. The key constraint is that the number must be divisible by 4. The divisibility rule for 4 is that the number formed by the last two digits (the tens and units place) must be divisible by 4. 
Step 2: Key Formula or Approach: 
This is a permutations and combinations problem. We will use a slot-based method. 
1. First, identify all possible pairs of digits from the given set that can form the last two digits of the number, satisfying the divisibility rule for 4. 
2. For each valid pair of last two digits, determine the number of available digits for the first two places. 
3. Use the permutation formula \(^nP_r = \frac{n!}{(n-r)!}\) to calculate the number of ways to arrange the remaining digits in the first two places. 
4. The total count will be the product of the number of valid endings and the number of ways to form the beginning of the number. 
Step 3: Detailed Explanation: 
Part A: Find all possible valid endings 
We need to find two-digit numbers formed from \(\{1, 2, 3, 4, 5, 6\}\) (without repetition) that are divisible by 4. Let's list them systematically: 
- Ending in 2: 12, 32, 52 (3 pairs)
- Ending in 4: 24, 64 (2 pairs)
- Ending in 6: 16, 36, 56 (3 pairs)
Total number of valid pairs for the last two digits = \(3 + 2 + 3 = 8\) pairs. 
These pairs are: \(\{12, 16, 24, 32, 36, 52, 56, 64\}.\) 
Part B: Fill the remaining places 
Let's consider any one of these 8 valid endings, for example, '12'. 
The last two digits are fixed. We have used the digits 1 and 2. 
The remaining available digits are \(\{3, 4, 5, 6\}\). There are 4 digits left. 
We need to fill the first two places (thousands and hundreds) of the 4-digit number. 
- The thousands place can be filled in any of the 4 remaining ways. 
- After filling the thousands place, the hundreds place can be filled in any of the remaining 3 ways. 
So, for each valid ending, the number of ways to fill the first two places is \(4 \times 3 = 12\) ways. (This is also \(^4P_2\)). 
Part C: Calculate the total number 
The total number of 4-digit numbers is the product of the number of valid endings and the number of ways to fill the beginning. 
Total Numbers = (Number of valid endings) \(\times\) (Ways to fill the first two places) 
Total Numbers = \(8 \times 12 = 96\). 
Step 4: Final Answer: 
The number of 4-digit numbers that can be formed is 96.