<p>Let <i>n</i> be an odd natural number greater than 1. Then, find the number of zeros at the end of the sum \(99^n + 1\).</p>
Step-by-Step Solution
Key Concept: Since n is odd, we can use the factorization a^n + b^n = (a+b)(a^(n-1) - a^(n-2)b + ... + b^(n-1)) when n is odd. Here 99^n + 1 = (99+1)(99^(n-1) - 99^(n-2) + ... + 1), and we must find the exact power of 10 dividing this product.
<p><strong>Step 1:</strong> Since n is odd and greater than 1, use the factorization: 99^n + 1^n = (99 + 1)(99^(n-1) - 99^(n-2) + 99^(n-3) - ... + 1)</p><p><strong>Step 2:</strong> Calculate the first factor: 99 + 1 = 100 = 10^2 = 2^2 × 5^2</p><p><strong>Step 3:</strong> For the second factor S = 99^(n-1) - 99^(n-2) + 99^(n-3) - ... + 1, note that it contains n terms (where n is odd). Since 99 ≡ -1 (mod 2), the alternating sum has n terms of alternating parity. With n odd, S ≡ 1 - 1 + 1 - ... + 1 ≡ 1 (mod 2), so S is odd and contributes no factors of 2.</p><p><strong>Step 4:</strong> Check divisibility by 5: S = (99^n + 1)/(99 + 1) and 99 ≡ 4 (mod 5). The alternating sum modulo 5 with n odd terms gives S ≢ 0 (mod 5).</p><p><strong>Step 5:</strong> The total power of 10 dividing 99^n + 1 is min(power of 2, power of 5) = min(2, 0) = 0 from the second factor. Only the first factor 100 = 10^2 contributes trailing zeros.</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2