Binomial Theorem
Divisibility using Mathematical Induction
Grade None
Question:
<p>If \(10^n + 3 \cdot 4^{n+2} + k\) is divisible by 9 for all \(n \in \mathbb{N}\), then the least positive integral value of \(k\) is __________.</p>
Step-by-Step Solution
Key Concept: Use the Binomial Theorem to express 10^n and 4^(n+2) in forms involving 9, then find the remainder condition. Since the expression must be divisible by 9 for ALL n ∈ ℕ, the remainder modulo 9 must be zero for every value of n.
<p><strong>Step 1:</strong> Express using Binomial Theorem. Write 10^n = (9+1)^n and 4^(n+2) = 16·4^n = 16(9-5)^n</p><p><strong>Step 2:</strong> Using Binomial expansion: 10^n ≡ 1 (mod 9) and 4^n ≡ (-5)^n (mod 9)</p><p><strong>Step 3:</strong> So 4^(n+2) = 16·4^n ≡ 16·(-5)^n ≡ 7·(-5)^n (mod 9)</p><p><strong>Step 4:</strong> The expression becomes: 1 + 3·16·(-5)^n + k ≡ 1 + 48·(-5)^n + k ≡ 1 + 3·(-5)^n + k (mod 9)</p><p><strong>Step 5:</strong> For n=1 (odd): 1 + 3(-5) + k ≡ 1 - 15 + k ≡ -14 + k ≡ k - 5 (mod 9). Setting k - 5 ≡ 0 gives k ≡ 5 (mod 9)</p><p><strong>Step 6:</strong> For n=2 (even): 1 + 3(25) + k ≡ 1 + 75 + k ≡ 76 + k ≡ 4 + k (mod 9). Setting 4 + k ≡ 0 gives k ≡ 5 (mod 9)</p><p><strong>Step 7:</strong> Both conditions consistently give k ≡ 5 (mod 9). The least positive integral value is k = 5.</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: 5