<p>How many five-digit numbers divisible by 6 can be made with the digits 0, 1, 2, 3, 4 and 5 if the digits cannot be repeated in the same number?</p>
Step-by-Step Solution
Key Concept: A number is divisible by 6 iff it's divisible by both 2 and 3. For divisibility by 3, the sum of digits must be divisible by 3. Since 0+1+2+3+4+5=15, we must exclude exactly one digit whose value ≡ 0 (mod 3) to maintain divisibility by 3, then ensure the last digit is even.
<p><strong>Step 1: Divisibility Conditions</strong></p><p>For divisibility by 6: must be divisible by both 2 and 3.</p><p><strong>Step 2: Check divisibility by 3</strong></p><p>Sum of all 6 digits: 0+1+2+3+4+5 = 15 ≡ 0 (mod 3). For any 5-digit selection, sum ≡ (value of excluded digit) (mod 3). To make sum ≡ 0 (mod 3), exclude digits from {0, 3} (digits ≡ 0 mod 3) or exclude digits from {1, 4} or {2, 5}.</p><p>Viable 5-digit sets:</p><p>• Exclude 0: {1,2,3,4,5}, sum = 15 ✓</p><p>• Exclude 3: {0,1,2,4,5}, sum = 12 ✓</p><p>• Exclude 1: {0,2,3,4,5}, sum = 14 ✗</p><p>• Exclude 4: {0,1,2,3,5}, sum = 11 ✗</p><p>• Exclude 2: {0,1,3,4,5}, sum = 13 ✗</p><p>• Exclude 5: {0,1,2,3,4}, sum = 10 ✗</p><p><strong>Step 3: Case 1 - Set {1,2,3,4,5}</strong></p><p>Last digit must be even: {2, 4}</p><p>• If last digit is 2: arrange {1,3,4,5} in first 4 positions = 4! = 24</p><p>• If last digit is 4: arrange {1,2,3,5} in first 4 positions = 4! = 24</p><p>Subtotal: 48</p><p><strong>Step 4: Case 2 - Set {0,1,2,4,5}</strong></p><p>Last digit must be even and non-zero: {2, 4}</p><p>• If last digit is 2: First digit from {1,4,5}, then arrange remaining 3 in middle 3 positions = 3 × 3! = 18</p><p>• If last digit is 4: First digit from {1,2,5}, then arrange remaining 3 in middle 3 positions = 3 × 3! = 18</p><p>Subtotal: 36</p><p><strong>Step 5: Total</strong></p><p>∴ Answer: 48 + 36 = <strong>84</strong></p>
Correct Answer: 48