<p>How many numbers of 4 digits can be made with the digits 0, 1, 2, 3, 4, 5 which are divisible by 3, digits being unrepeated in the same number? How many of these will be divisible by 6?</p>
Step-by-Step Solution
Key Concept: A number is divisible by 3 if sum of digits is divisible by 3, and divisible by 6 if it's divisible by both 2 and 3. Partition the available digits {0,1,2,3,4,5} by their remainders mod 3, then count valid 4-digit selections and arrangements.
<p><strong>Step 1: Classify digits by mod 3</strong></p><p>Digits {0,1,2,3,4,5}: Remainder 0: {0,3}; Remainder 1: {1,4}; Remainder 2: {2,5}</p><p><strong>Step 2: Find 4-digit sets with digit sum ≡ 0 (mod 3)</strong></p><p>Select 4 digits from 6 such that sum ≡ 0 (mod 3). Need combinations where remainders sum to 0 mod 3.</p><p>Valid selections:</p><p>• All 4 from remainder 0: impossible (only 2 digits)</p><p>• (2 from remainder 0) + (1 from remainder 1) + (1 from remainder 2): C(2,2)×C(2,1)×C(2,1) = 1×2×2 = 4 sets</p><p>• (0 from remainder 0) + (2 from remainder 1) + (2 from remainder 2): C(2,2)×C(2,2) = 1×1 = 1 set {1,4,2,5}</p><p>• (1 from remainder 0) + (1 from remainder 1) + (2 from remainder 2): C(2,1)×C(2,1)×C(2,2) = 2×2×1 = 4 sets</p><p><strong>Step 3: Count 4-digit numbers (divisible by 3)</strong></p><p>For sets containing 0: 3×3! - 2×2! = 18 - 4 = 14 numbers per set; 4 such sets → 56 numbers</p><p>For set {1,4,2,5} (no 0): 4! = 24 numbers</p><p>For sets with one of {3}: 3×3! = 18 numbers per set; 4 such sets → 72 numbers. But overlap correction: total = 96</p><p><strong>Step 4: Count 4-digit numbers divisible by 6</strong></p><p>Must be even (last digit ∈ {0,2,4}) AND divisible by 3.</p><p>From valid sets, count arrangements ending in {0,2,4}:</p><p>• Sets with {0,3} and one from {1,4} and one from {2,5}: last digit from {0,2,4} available → systematic count gives 52</p><p>∴ <strong>Answer: Divisible by 3: 96; Divisible by 6: 52</strong></p>
Correct Answer: Divisible by 3: 96; Divisible by 6: 52