Key Concepts and Formulas

• A number is divisible by 4 if its last two digits are divisible by 4.
• The number of ways to arrange $n$ distinct objects in $r$ places is given by the permutation formula $P(n,r) = \frac{n!}{(n-r)!}$. This is equivalent to $n \times (n-1) \times \dots \times (n-r+1)$.

Step-by-Step Solution

Step 1: Define the range and divisibility rule.

We need to form numbers between 1000 and 3000 using the digits 1, 2, 3, 4, 5, and 6 without repetition, such that the numbers are divisible by 4. A number is divisible by 4 if its last two digits are divisible by 4.

Step 2: Determine the possible first digits.

Since the numbers must be between 1000 and 3000, the first digit can only be 1 or 2.

Step 3: Case 1: The first digit is 1.

If the first digit is 1, we need to find the possible combinations for the last two digits such that they form a number divisible by 4. We consider the digits 1, 2, 3, 4, 5, and 6. The last two digits must be divisible by 4. The possible pairs are: 24, 32, 36, 52, 56, 64.

Step 4: Calculate the number of possibilities for Case 1.

Since the first digit is 1 and the last two digits are chosen, we have one digit fixed in the thousands place and two digits fixed in the tens and units places. We have 6 digits to choose from. We have used 3 digits already. So, we have $6 - 3 = 3$ digits remaining to choose from for the hundreds place.

The number of such numbers in Case 1 is $1 \times 3 \times 6 = 18$. This is because we fix the first digit as 1, then we have 6 choices for the last two digits (24, 32, 36, 52, 56, 64), and then we have 3 choices for the hundreds place (since we cannot repeat digits).

Step 5: Case 2: The first digit is 2.

If the first digit is 2, we need to find the possible combinations for the last two digits such that they form a number divisible by 4. The possible pairs are: 16, 36, 56, 64. Note that 24, 32, 52, 12 are divisible by 4, but the digit 2 is already used in the thousand's place.

Step 6: Calculate the number of possibilities for Case 2.

Since the first digit is 2 and the last two digits are chosen, we have one digit fixed in the thousands place and two digits fixed in the tens and units places. We have 6 digits to choose from. We have used 3 digits already. So, we have $6 - 3 = 3$ digits remaining to choose from for the hundreds place.

The number of such numbers in Case 2 is $1 \times 3 \times 4 = 12$. This is because we fix the first digit as 2, then we have 4 choices for the last two digits (16, 36, 56, 64), and then we have 3 choices for the hundreds place (since we cannot repeat digits).

Step 7: Calculate the total number of such numbers.

The total number of such numbers is the sum of the possibilities from Case 1 and Case 2: $18 + 12 = 30$.

Common Mistakes & Tips

• Remember to check if the chosen digits are already used in other positions.
• Make sure the last two digits are divisible by 4.
• Don't forget that the first digit restricts the range of the numbers.

Summary

We considered the two possible cases for the first digit (1 and 2) and then found the possible combinations for the last two digits to ensure divisibility by 4. We then calculated the number of possibilities for the hundreds place in each case and summed the results to find the total number of such numbers.

Final Answer

The final answer is \boxed{30}.