<p>Find the number of three-digit numbers in which repetition is allowed and sum of digits is even.</p>
Step-by-Step Solution
Key Concept: A sum is even when we have either all even digits or exactly two odd digits. Use the complementary counting approach: total three-digit numbers with even digit sum = (total with even sum in last two digits) × (all choices for first digit) organized by parity.
<p><strong>Step 1:</strong> For a three-digit number, first digit ∈ {1,2,...,9} (9 choices), and second and third digits ∈ {0,1,...,9} (10 choices each).</p><p><strong>Step 2:</strong> For sum of digits to be even, we need: (first digit + second digit + third digit) is even. This happens when the digits contain an even number of odd digits (0 or 2 odd digits).</p><p><strong>Step 3:</strong> <strong>Case 1 - All three digits even:</strong> First digit even ∈ {2,4,6,8} = 4 choices. Second and third digits even ∈ {0,2,4,6,8} = 5 choices each. Count = 4 × 5 × 5 = 100</p><p><strong>Step 4:</strong> <strong>Case 2 - Exactly two digits odd:</strong> We choose which 2 of 3 positions have odd digits.</p><p>• <strong>Subcase 2a:</strong> First and second odd, third even: First digit odd ∈ {1,3,5,7,9} = 5 choices. Second digit odd ∈ {1,3,5,7,9} = 5 choices. Third digit even ∈ {0,2,4,6,8} = 5 choices. Count = 5 × 5 × 5 = 125</p><p>• <strong>Subcase 2b:</strong> First and third odd, second even: First digit odd = 5 choices. Second digit even = 5 choices. Third digit odd = 5 choices. Count = 5 × 5 × 5 = 125</p><p>• <strong>Subcase 2c:</strong> Second and third odd, first even: First digit even (but ≠0) ∈ {2,4,6,8} = 4 choices. Second digit odd = 5 choices. Third digit odd = 5 choices. Count = 4 × 5 × 5 = 100</p><p><strong>Step 5:</strong> Total = 100 + 125 + 125 + 100 = <strong>450</strong></p>
Correct Answer: 450