<p>How many 4-digit numbers between 2000 and 5000 are multiples of 3, formed using the digits 0, 1, 2, 3, 4 (repetition not allowed)?</p>
Step-by-Step Solution
Key Concept: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. First identify valid digit sets, then count valid arrangements with first digit constraint (2, 3, or 4).
<p><strong>Step 1: Find digit sum constraint</strong></p><p>Sum of all available digits: 0+1+2+3+4 = 10. We need 4 digits whose sum ≡ 0 (mod 3).</p><p><strong>Step 2: Identify valid 4-digit sets</strong></p><p>Removing one digit at a time:<br>• Remove 0: {1,2,3,4}, sum = 10 ≡ 1 (mod 3) ✗<br>• Remove 1: {0,2,3,4}, sum = 9 ≡ 0 (mod 3) ✓<br>• Remove 2: {0,1,3,4}, sum = 8 ≡ 2 (mod 3) ✗<br>• Remove 3: {0,1,2,4}, sum = 7 ≡ 1 (mod 3) ✗<br>• Remove 4: {0,1,2,3}, sum = 6 ≡ 0 (mod 3) ✓</p><p><strong>Step 3: Count valid arrangements for {0,2,3,4}</strong></p><p>Valid first digits (between 2000-5000): 2, 3, or 4<br>• First digit = 2: arrange {0,3,4} → 3! = 6 ways<br>• First digit = 3: arrange {0,2,4} → 3! = 6 ways<br>• First digit = 4: arrange {0,2,3} → 3! = 6 ways<br>Subtotal: 18 ways</p><p><strong>Step 4: Count valid arrangements for {0,1,2,3}</strong></p><p>Valid first digits: 2 or 3 (4 is not available)<br>• First digit = 2: arrange {0,1,3} → 3! = 6 ways<br>• First digit = 3: arrange {0,1,2} → 3! = 6 ways<br>Subtotal: 12 ways</p><p><strong>Step 5: Total</strong></p><p>18 + 12 = <strong>30</strong></p><p>∴ Answer: B</p>
Correct Answer: B