<p>\(n\) is selected from the set \(\{1, 2, 3, \ldots, 10\}\) and the number \(2^n + 3^n + 5^n\) is formed. Total number of ways of selecting \(n\) so that the formed number is divisible by 4 is equal to</p>
Step-by-Step Solution
Key Concept: For divisibility by 4, we need 2^n + 3^n + 5^n ≡ 0 (mod 4). Since 2^n ≡ 0 (mod 4) for n ≥ 2, we only need 3^n + 5^n ≡ 0 (mod 4). Both 3 and 5 are odd, so their powers alternate between odd patterns modulo 4.
<p><strong>Step 1:</strong> Analyze 2^n + 3^n + 5^n (mod 4) for divisibility by 4.</p><p><strong>Step 2:</strong> For n = 1: 2¹ + 3¹ + 5¹ = 2 + 3 + 5 = 10 ≡ 2 (mod 4) ✗</p><p><strong>Step 3:</strong> For n ≥ 2: 2^n ≡ 0 (mod 4), so we need 3^n + 5^n ≡ 0 (mod 4).</p><p><strong>Step 4:</strong> Check pattern of 3^n (mod 4): 3¹ ≡ 3, 3² ≡ 1, 3³ ≡ 3, 3⁴ ≡ 1,... (odd n gives 3, even n gives 1)</p><p><strong>Step 5:</strong> Check pattern of 5^n (mod 4): 5¹ ≡ 1, 5² ≡ 1, 5³ ≡ 1,... (always ≡ 1)</p><p><strong>Step 6:</strong> For 3^n + 5^n ≡ 0 (mod 4):</p><ul><li>If n is odd: 3^n + 5^n ≡ 3 + 1 ≡ 0 (mod 4) ✓</li><li>If n is even: 3^n + 5^n ≡ 1 + 1 ≡ 2 (mod 4) ✗</li></ul><p><strong>Step 7:</strong> From {1,2,3,...,10}, odd values ≥ 2 are: {3, 5, 7, 9}</p><p>∴ Answer: 4 (which is option D)
Correct Answer: D