Step-by-Step Solution
Key Concept: To find the last digit of a sum, we need the last digit of each term separately. The last digit of 9! is 0 (since 9! contains factors 2 and 5), and we need to find the pattern of last digits of powers of 3.
<p><strong>Step 1: Find the last digit of 9!</strong></p><p>9! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 362880</p><p>Since 9! contains both 2 and 5 as factors, it's divisible by 10, so the last digit of 9! is <strong>0</strong>.</p><p><strong>Step 2: Find the pattern of last digits of powers of 3</strong></p><p>3¹ = 3 (last digit: 3)</p><p>3² = 9 (last digit: 9)</p><p>3³ = 27 (last digit: 7)</p><p>3⁴ = 81 (last digit: 1)</p><p>3⁵ = 243 (last digit: 3)</p><p>The pattern repeats every 4 powers: {3, 9, 7, 1}</p><p><strong>Step 3: Find the position of 3⁹⁹⁶⁶ in the cycle</strong></p><p>Divide the exponent by the cycle length: 9966 ÷ 4 = 2491 remainder 2</p><p>Since the remainder is 2, 3⁹⁹⁶⁶ has the same last digit as 3² = 9</p><p><strong>Step 4: Find the last digit of the sum</strong></p><p>Last digit of (9! + 3⁹⁹⁶⁶) = Last digit of (0 + 9) = <strong>9</strong></p><p>∴ Answer: D</p>
Correct Answer: D