<p><strong>Q27.</strong> Let us define D(n) = sum of all the digits of n (Digital sum of number n). So D(10) = 1 + 0 = 1, D(7896) = 7 + 8 + 9 + 6 = 30. Then find \(D(D(D(4444^{4444})))\).</p>
Step-by-Step Solution
Key Concept: Use the property that a number and its digital sum are congruent mod 9 (since 10 ≡ 1 mod 9). Therefore D(n) ≡ n (mod 9), so we only need to find 4444^4444 mod 9 to determine the final answer.
<p><strong>Step 1: Find 4444 mod 9</strong></p><p>4444 = 4+4+4+4 = 16 ≡ 1+6 = 7 (mod 9)</p><p><strong>Step 2: Find 4444^4444 mod 9</strong></p><p>Since 4444 ≡ 7 (mod 9), we need 7^4444 (mod 9)</p><p>Powers of 7 mod 9: 7¹ ≡ 7, 7² ≡ 49 ≡ 4, 7³ ≡ 28 ≡ 1 (mod 9)</p><p>The order is 3, so 7^4444 ≡ 7^(4444 mod 3) (mod 9)</p><p>4444 = 3(1481) + 1, so 4444 ≡ 1 (mod 3)</p><p>Therefore 7^4444 ≡ 7¹ ≡ 7 (mod 9)</p><p><strong>Step 3: Bound the digital sum</strong></p><p>4444^4444 has at most 4444 × log₁₀(4444) ≈ 4444 × 3.65 ≈ 16,200 digits</p><p>So D(4444^4444) ≤ 9 × 16,200 = 145,800</p><p>Then D(D(4444^4444)) ≤ 9 × 6 = 54</p><p>Then D(D(D(4444^4444))) ≤ 9 + 9 = 18 (single or two digits)</p><p><strong>Step 4: Apply mod 9 constraint</strong></p><p>Since D(D(D(4444^4444))) ≡ 7 (mod 9) and it's a single or two-digit number ≤ 18:</p><p>The only value ≡ 7 (mod 9) in range [1,18] is 7 and 16</p><p>Since D(D(D(n))) for n ≡ 7 (mod 9) must be in {7, 16}, and 16 has digit sum 7, the stable fixed point is 7.</p><p>∴ <strong>Answer: 7</strong></p>
Correct Answer: 7