<p>Two non-negative integers are chosen at random from the set of non-negative integers with replacement. What is the probability that the sum of their squares is divisible by 10?</p>
Step-by-Step Solution
Key Concept: Analyze the last digit of perfect squares (which depend only on the last digit of the base number): squares can only end in 0,1,4,5,6,9. For sum of two squares to be divisible by 10, we need the sum of last digits to end in 0.
<p><strong>Step 1:</strong> Determine which last digits are possible for perfect squares mod 10.</p><p>Checking n² mod 10 for n = 0,1,2,...,9: we get {0,1,4,9,6,5,6,9,4,1}.</p><p>The possible last digits of perfect squares are: <strong>{0,1,4,5,6,9}</strong> (6 residues out of 10).</p><p><strong>Step 2:</strong> Count pairs (a,b) where a² + b² ≡ 0 (mod 10).</p><p>If we choose the last digit of a and b uniformly from {0,1,4,5,6,9}, we need a² + b² to end in 0.</p><p>Valid pairs from last digits: (0,0), (1,9), (9,1), (4,6), (6,4), (5,5).</p><p>That's 6 valid pairs out of 6×6 = 36 possible pairs.</p><p><strong>Step 3:</strong> Calculate probability.</p><p>In the limit of choosing from non-negative integers, the probability that the last digit of each number is uniformly distributed over {0,1,2,...,9} is 1/10 for each digit.</p><p>P(last digit of a ∈ S) = 6/10 for S = {0,1,4,5,6,9}.</p><p>P(a² + b² ≡ 0 mod 10) = (6/10) × (6/10) × (6/36) = 36/100 × 1/6 = <strong>6/100 = 3/50</strong></p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C