1. **Problem Statement:** We need to find the total number of 4-digit numbers between 1000 and 3000 that are divisible by 4, formed using the digits 1, 2, 3, 4, 5, and 6 without repeating any digit. 2. **Key Points:** - The number is 4-digit, so the first digit can be 1 or 2 (since the number must be between 1000 and 3000). - The digits cannot repeat. - The number must be divisible by 4. 3. **Divisibility rule for 4:** A number is divisible by 4 if the number formed by its last two digits is divisible by 4. 4. **Step 1: Determine possible first digits:** - First digit $d_1$ can be 1 or 2. 5. **Step 2: Find all possible last two digits $d_3 d_4$ that are divisible by 4, using digits from {1,2,3,4,5,6} without repetition and not repeating the first digit.** - List all two-digit numbers divisible by 4 from digits 1 to 6 without repetition: - 12 (12 % 4 = 0) - 16 (16 % 4 = 0) - 24 (24 % 4 = 0) - 32 (32 % 4 = 0) - 36 (36 % 4 = 0) - 44 (44 % 4 = 0) but digit 4 repeats, so invalid - 52 (52 % 4 = 0) - 56 (56 % 4 = 0) - 64 (64 % 4 = 0) - Valid last two digits pairs: 12, 16, 24, 32, 36, 52, 56, 64 6. **Step 3: For each first digit, find valid last two digits that do not repeat the first digit:** - If $d_1=1$: - Last two digits cannot contain 1. - From the list: 12 (contains 1), 16 (contains 1), 24, 32, 36, 52, 56, 64 - Remove 12 and 16. - Valid last two digits: 24, 32, 36, 52, 56, 64 - If $d_1=2$: - Last two digits cannot contain 2. - From the list: 12 (contains 2), 16, 24 (contains 2), 32 (contains 2), 36, 52 (contains 2), 56, 64 - Remove 12, 24, 32, 52. - Valid last two digits: 16, 36, 56, 64 7. **Step 4: For each valid combination of $d_1$ and last two digits, find possible $d_2$ (second digit) from remaining digits without repetition:** - Total digits: {1,2,3,4,5,6} - Case $d_1=1$: - Valid last two digits: 24, 32, 36, 52, 56, 64 - For each, remove digits used in $d_1$ and last two digits to find $d_2$ options. - Example: last two digits = 24 - Used digits: 1,2,4 - Remaining digits for $d_2$: {3,5,6} (3 digits) - Similarly for others: - 32: used {1,3,2}, remaining {4,5,6} (3 digits) - 36: used {1,3,6}, remaining {2,4,5} (3 digits) - 52: used {1,5,2}, remaining {3,4,6} (3 digits) - 56: used {1,5,6}, remaining {2,3,4} (3 digits) - 64: used {1,6,4}, remaining {2,3,5} (3 digits) - For each last two digits, 3 choices for $d_2$. - Number of valid numbers for $d_1=1$ is $6 \times 3 = 18$. - Case $d_1=2$: - Valid last two digits: 16, 36, 56, 64 - For each, remove digits used in $d_1$ and last two digits to find $d_2$ options. - 16: used {2,1,6}, remaining {3,4,5} (3 digits) - 36: used {2,3,6}, remaining {1,4,5} (3 digits) - 56: used {2,5,6}, remaining {1,3,4} (3 digits) - 64: used {2,6,4}, remaining {1,3,5} (3 digits) - For each last two digits, 3 choices for $d_2$. - Number of valid numbers for $d_1=2$ is $4 \times 3 = 12$. 8. **Step 5: Total numbers = numbers with $d_1=1$ + numbers with $d_1=2$ = 18 + 12 = 30$. **Final answer:** $$\boxed{30}$$