<p>Two natural numbers <em>x</em> and <em>y</em> are chosen at random. What is the probability that <em>x</em><sup>2</sup> + <em>y</em><sup>2</sup> is divisible by 5?</p>
Step-by-Step Solution
Key Concept: Analyze quadratic residues modulo 5: x² and y² can only be ≡ 0, 1, or 4 (mod 5). For x² + y² ≡ 0 (mod 5), we need specific combinations of these residues.
<p><strong>Step 1:</strong> Find all possible values of x² mod 5 and y² mod 5.</p><p>For any integer n: if n ≡ 0 (mod 5), then n² ≡ 0 (mod 5); if n ≡ ±1 (mod 5), then n² ≡ 1 (mod 5); if n ≡ ±2 (mod 5), then n² ≡ 4 (mod 5).</p><p>So x² mod 5 ∈ {0, 1, 4} and y² mod 5 ∈ {0, 1, 4}.</p><p><strong>Step 2:</strong> Determine when x ≡ 0, 1, 2, 3, 4 (mod 5).</p><p>Out of every 5 consecutive natural numbers: exactly 1 gives x² ≡ 0 (mod 5), exactly 2 give x² ≡ 1 (mod 5), and exactly 2 give x² ≡ 4 (mod 5).</p><p>So P(x² ≡ 0) = 1/5, P(x² ≡ 1) = 2/5, P(x² ≡ 4) = 2/5. Same for y.</p><p><strong>Step 3:</strong> Find when x² + y² ≡ 0 (mod 5).</p><p>Possible combinations: (0,0), (1,4), (4,1).</p><p>P(x² + y² ≡ 0 mod 5) = P(x²≡0)·P(y²≡0) + P(x²≡1)·P(y²≡4) + P(x²≡4)·P(y²≡1)</p><p>= (1/5)(1/5) + (2/5)(2/5) + (2/5)(2/5)</p><p>= 1/25 + 4/25 + 4/25 = 9/25</p><p>∴ Answer: B (9/25)</p>
Correct Answer: B