<p>Two numbers <em>x</em> and <em>y</em> are chosen at random from the set {1, 2, 3, ..., 3n}. Find the probability that <em>x</em><sup>2</sup> − <em>y</em><sup>2</sup> is divisible by 3.</p>
<p>\(\dfrac{3n-3}{3(3n-1)}\)</p>
<p>\(\dfrac{5n-3}{3(3n-1)}\)</p>
<p>\(\dfrac{3n+3}{3(3n-1)}\)</p>
<p>\(\dfrac{5n+3}{3(3n-1)}\)</p>
Step-by-Step Solution
Key Concept: x² − y² ≡ 0 (mod 3) requires x² ≡ y² (mod 3). Since any integer squared is congruent to either 0 or 1 (mod 3), we need both x and y in the same residue class modulo 3.
<p><strong>Step 1: Analyze x² mod 3</strong></p><p>For any integer a: a ≡ 0, 1, or 2 (mod 3)</p><p>• If a ≡ 0 (mod 3), then a² ≡ 0 (mod 3)</p><p>• If a ≡ 1 (mod 3), then a² ≡ 1 (mod 3)</p><p>• If a ≡ 2 (mod 3), then a² ≡ 4 ≡ 1 (mod 3)</p><p>So x² ≡ 0 or 1 (mod 3) only.</p><p><strong>Step 2: Condition for divisibility</strong></p><p>x² − y² ≡ 0 (mod 3) requires x² ≡ y² (mod 3)</p><p>This means both must be ≡ 0 (mod 3) OR both must be ≡ 1 (mod 3)</p><p><strong>Step 3: Count numbers by residue class</strong></p><p>In {1, 2, ..., 3n}:</p><p>• Numbers ≡ 0 (mod 3): {3, 6, 9, ..., 3n} → n numbers</p><p>• Numbers ≡ 1 (mod 3): {1, 4, 7, ..., 3n-2} → n numbers</p><p>• Numbers ≡ 2 (mod 3): {2, 5, 8, ..., 3n-1} → n numbers</p><p><strong>Step 4: Count favorable outcomes</strong></p><p>Case 1: Both x, y ≡ 0 (mod 3) and x ≠ y: n(n-1) ways</p><p>Case 2: Both x, y ≡ 1 (mod 3) and x ≠ y: n(n-1) ways</p><p>Case 3: Both x, y ≡ 2 (mod 3) and x ≠ y: n(n-1) ways</p><p>Total favorable = 3n(n-1)</p><p><strong>Step 5: Calculate probability</strong></p><p>Total outcomes = 3n(3n-1)</p><p>Probability = 3n(n-1)/[3n(3n-1)] = (n-1)/(3n-1)</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B