<p>The last digit of \((1! + 2! + \ldots + 2005!)^{500}\) is</p>
Step-by-Step Solution
Key Concept: For n ≥ 5, n! ends in 0 (divisible by 10), so the sum 1! + 2! + ... + 2005! has the same last digit as 1! + 2! + 3! + 4!. Then find the last digit of this sum raised to power 500.
<p><strong>Step 1:</strong> Recognize that for all n ≥ 5, n! is divisible by 10 (contains factors 2 and 5), so n! ends in 0.</p><p><strong>Step 2:</strong> Calculate 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33. The last digit is 3.</p><p><strong>Step 3:</strong> Since all terms from 5! onwards end in 0, the sum 1! + 2! + ... + 2005! ends in 3.</p><p><strong>Step 4:</strong> Find the last digit of 3^500. The pattern of last digits of powers of 3 is: 3¹=3, 3²=9, 3³=7, 3⁴=1, 3⁵=3, ... (repeats with period 4).</p><p><strong>Step 5:</strong> Since 500 = 4 × 125, we have 500 ≡ 0 (mod 4). Therefore 3^500 has the same last digit as 3⁴, which is 1.</p><p>∴ Answer: D (last digit is 1)</p>
Correct Answer: D