<p>Let <em>P(n)</em> = 10<sup>n</sup> + 3·4<sup>n+2</sup> + k is divisible by 9, ∀n ∈ N. The least positive integral value of k is:</p>
Step-by-Step Solution
Key Concept: Use mathematical induction or direct divisibility analysis: rewrite 4^(n+2) = 16·4^n, then use modular arithmetic (mod 9) to find the remainder when 10^n + 48·4^n is divided by 9 for various n, and determine k that makes the entire expression ≡ 0 (mod 9).
<p><strong>Step 1:</strong> Rewrite P(n) = 10^n + 3·4^(n+2) + k = 10^n + 3·16·4^n + k = 10^n + 48·4^n + k</p><p><strong>Step 2:</strong> Find patterns modulo 9: Since 10 ≡ 1 (mod 9), we have 10^n ≡ 1 (mod 9) for all n ∈ ℕ.</p><p><strong>Step 3:</strong> Find 4^n (mod 9) pattern: 4^1 ≡ 4, 4^2 ≡ 16 ≡ 7, 4^3 ≡ 28 ≡ 1 (mod 9). The cycle repeats with period 3.</p><p><strong>Step 4:</strong> Calculate 48·4^n (mod 9): Since 48 ≡ 3 (mod 9), we get 48·4^n ≡ 3·4^n (mod 9).</p><p><strong>Step 5:</strong> For n ≡ 1 (mod 3): P(n) ≡ 1 + 3(4) + k ≡ 1 + 12 + k ≡ 13 + k ≡ 4 + k (mod 9)</p><p><strong>Step 6:</strong> For n ≡ 2 (mod 3): P(n) ≡ 1 + 3(7) + k ≡ 1 + 21 + k ≡ 22 + k ≡ 4 + k (mod 9)</p><p><strong>Step 7:</strong> For n ≡ 0 (mod 3): P(n) ≡ 1 + 3(1) + k ≡ 1 + 3 + k ≡ 4 + k (mod 9)</p><p><strong>Step 8:</strong> All cases give 4 + k ≡ 0 (mod 9), so k ≡ 5 (mod 9). The least positive integral value is <strong>k = 5</strong>.</p><p>∴ Answer: D</p>
Correct Answer: D