4 Digit Divisible 4 A8401B

1. **Problem Statement:** Find the number of 4-digit numbers that do not include the digits 7 and 8, are divisible by 4, and have no repeated digits. 2. **Key Points:** - The digits allowed are from 0-9 except 7 and 8, so digits are \{0,1,2,3,4,5,6,9\}. - The number is 4-digit, so the first digit cannot be 0. - The number must be divisible by 4. - No digit repeats. 3. **Divisibility rule for 4:** A number is divisible by 4 if its last two digits form a number divisible by 4. 4. **Step 1: Identify possible last two digits (\(d_3 d_4\)) divisible by 4 from allowed digits without repetition.** - Allowed digits for last two positions: \{0,1,2,3,4,5,6,9\} - List all two-digit numbers from these digits divisible by 4: - 00 (digits repeat, so invalid) - 12 (12 % 4 = 0) - 16 (16 % 4 = 0) - 20 (20 % 4 = 0) - 24 (24 % 4 = 0) - 32 (32 % 4 = 0) - 36 (36 % 4 = 0) - 40 (40 % 4 = 0) - 44 (digits repeat, invalid) - 52 (52 % 4 = 0) - 56 (56 % 4 = 0) - 60 (60 % 4 = 0) - 64 (64 % 4 = 0) - 92 (92 % 4 = 0) - 96 (96 % 4 = 0) - Remove pairs with repeated digits (like 00, 44) and digits 7 or 8. - Valid last two digits pairs: 12,16,20,24,32,36,40,52,56,60,64,92,96 5. **Step 2: For each valid last two digits pair, count possible first two digits (\(d_1 d_2\))** - \(d_1\) (thousands place) can be any allowed digit except 0 and digits used in last two digits. - \(d_2\) (hundreds place) can be any allowed digit except digits used in \(d_1\) and last two digits. 6. **Calculate for each pair:** - Total allowed digits: 8 (0,1,2,3,4,5,6,9) - Exclude digits in last two digits (2 digits) - Exclude 0 for \(d_1\) Example for last two digits = 12: - Digits used: 1,2 - \(d_1\) choices: from \{3,4,5,6,9\} (excluding 0 and digits 1,2) = 5 choices - \(d_2\) choices: from remaining digits excluding \(d_1\) and last two digits - Total digits left: 8 - 2 (last two) - 1 (\(d_1\)) = 5 choices Number of numbers for last two digits 12 = 5 * 5 = 25 Repeat this for each pair: - 16: digits used 1,6; \(d_1\) from \{2,3,4,5,9\} = 5; \(d_2\) = 5; total 25 - 20: digits 2,0; \(d_1\) from \{1,3,4,5,6,9\} excluding 0 and 2 = 6; \(d_2\) = 5; total 30 - 24: digits 2,4; \(d_1\) from \{1,3,5,6,9\} = 5; \(d_2\) = 5; total 25 - 32: digits 3,2; \(d_1\) from \{1,4,5,6,9\} = 5; \(d_2\) = 5; total 25 - 36: digits 3,6; \(d_1\) from \{1,2,4,5,9\} = 5; \(d_2\) = 5; total 25 - 40: digits 4,0; \(d_1\) from \{1,2,3,5,6,9\} = 6; \(d_2\) = 5; total 30 - 52: digits 5,2; \(d_1\) from \{1,3,4,6,9\} = 5; \(d_2\) = 5; total 25 - 56: digits 5,6; \(d_1\) from \{1,2,3,4,9\} = 5; \(d_2\) = 5; total 25 - 60: digits 6,0; \(d_1\) from \{1,2,3,4,5,9\} = 6; \(d_2\) = 5; total 30 - 64: digits 6,4; \(d_1\) from \{1,2,3,5,9\} = 5; \(d_2\) = 5; total 25 - 92: digits 9,2; \(d_1\) from \{1,3,4,5,6\} = 5; \(d_2\) = 5; total 25 - 96: digits 9,6; \(d_1\) from \{1,2,3,4,5\} = 5; \(d_2\) = 5; total 25 7. **Step 3: Sum all totals:** - Totals: 25 + 25 + 30 + 25 + 25 + 25 + 30 + 25 + 25 + 30 + 25 + 25 + 25 - Sum = (25*9) + (30*4) = 225 + 120 = 345 **Final answer:** $$\boxed{345}$$ This is the number of 4-digit numbers without digits 7 and 8, divisible by 4, with no repeated digits.