<p>If \(10^m\) divides the number \(101^{100} - 1\), then find the greatest value of \(m\).</p>
Step-by-Step Solution
Key Concept: Use the Lifting the Exponent (LTE) lemma for odd primes: if p is odd, p|a-b, and p∤a, then vₚ(aⁿ - bⁿ) = vₚ(a - b) + vₚ(n). For p=2 with odd a,b and even n: v₂(aⁿ - bⁿ) = v₂(a - b) + v₂(a + b) + v₂(n) - 1. Then combine results using the Chinese Remainder Theorem concept for 10ᵐ = 2ᵐ · 5ᵐ.
<p><strong>Step 1: Factor using difference of powers</strong></p><p>101¹⁰⁰ - 1 = (101 - 1)(101⁹⁹ + 101⁹⁸ + ... + 101 + 1) = 100 · S, where S is the sum.</p><p><strong>Step 2: Find v₅(101¹⁰⁰ - 1) using LTE for p=5</strong></p><p>Since 5 | (101 - 1) = 100 and 5 ∤ 101:</p><p>v₅(101¹⁰⁰ - 1) = v₅(101 - 1) + v₅(100) = v₅(100) + v₅(100) = 2 + 2 = 4</p><p><strong>Step 3: Find v₂(101¹⁰⁰ - 1) using LTE for p=2</strong></p><p>Since 101 and 1 are both odd, and 100 is even:</p><p>v₂(101¹⁰⁰ - 1) = v₂(101 - 1) + v₂(101 + 1) + v₂(100) - 1</p><p>= v₂(100) + v₂(102) + v₂(100) - 1</p><p>= 2 + 1 + 2 - 1 = 4</p><p><strong>Step 4: Determine m</strong></p><p>For 10ᵐ = 2ᵐ · 5ᵐ to divide 101¹⁰⁰ - 1, we need m ≤ min(v₂, v₅) = min(4, 4) = 4</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4