1. **Problem Statement:** Find the total number $N$ of 5-digit numbers divisible by 4 formed using digits 1, 2, 3, 4, 5, and 6 without repetition. 2. **Key Rule for Divisibility by 4:** A number is divisible by 4 if its last two digits form a number divisible by 4. 3. **Step 1: Identify possible last two digits (units and tens) divisible by 4 from given digits without repetition:** Possible pairs from digits {1,2,3,4,5,6} are: - 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 repeated, so invalid - 52 (52 % 4 = 0) - 56 (56 % 4 = 0) - 64 (64 % 4 = 0) Valid last two digits without repetition are: 12, 16, 24, 32, 36, 52, 56, 64 4. **Step 2: For each valid last two digits, count the number of ways to choose the first three digits:** - Total digits: 6 - Last two digits fixed (2 digits used) - Remaining digits available: 6 - 2 = 4 - First digit cannot be zero (not in given digits, so no issue) - Number of ways to arrange first 3 digits from remaining 4 digits without repetition = $P(4,3) = 4 \times 3 \times 2 = 24$ 5. **Step 3: Calculate total numbers:** - Number of valid last two digit pairs = 8 - For each pair, 24 ways to arrange first three digits - Total $N = 8 \times 24 = 192$ **Final answer:** $$N = 192$$