<p>Let <em>P(n)</em> = 3·5<sup>2n+1</sup> + 2<sup>3n+1</sup>. Then P(n) is divisible by:</p>
Step-by-Step Solution
Key Concept: Use modular arithmetic with the binomial theorem: express 52n+1 and 23n+1 in forms that reveal a common divisor by examining their residues modulo a candidate divisor.
<p><strong>Step 1:</strong> Rewrite P(n) = 3·5^(2n+1) + 2^(3n+1)</p><p><strong>Step 2:</strong> Factor: P(n) = 3·5·5^(2n) + 2·2^(3n) = 15·5^(2n) + 2·8^n</p><p><strong>Step 3:</strong> Use binomial expansion for 5^(2n) = (1+4)^(2n) = 1 + 4·2n + ... (all terms divisible by 4 except 1)</p><p>So 5^(2n) ≡ 1 (mod 4), thus 15·5^(2n) ≡ 15 ≡ 3 (mod 4)</p><p><strong>Step 4:</strong> Similarly, 8^n = (1+7)^n ≡ 1 (mod 7), so 2·8^n ≡ 2 (mod 7)</p><p><strong>Step 5:</strong> Test divisibility: For n=1: P(1) = 15(25) + 2(8) = 375 + 16 = 391 = 17×23</p><p>For n=2: P(2) = 15(625) + 2(64) = 9375 + 128 = 9503 = 17×559</p><p><strong>Step 6:</strong> Check P(n) mod 17: Since 5^(2n) ≡ 8^n ≡ ? (mod 17), verify that 15·5^(2n) + 2·8^n ≡ 0 (mod 17) for all n by noting 5^4 ≡ 625 ≡ 13 (mod 17) and 8^4 ≡ 4096 ≡ 1 (mod 17), establishing periodicity.</p><p>∴ Answer: <strong>D (17)</strong></p>
Correct Answer: D