<p>Find the number of positive integral solutions of \(xyz = 21600\).</p>
Step-by-Step Solution
Key Concept: Factorize 21600 into prime factors, then distribute these prime powers among three variables x, y, z using the stars-and-bars method for each prime independently.
<p><strong>Step 1:</strong> Find prime factorization of 21600.</p><p>21600 = 216 × 100 = 6³ × 10² = (2·3)³ × (2·5)² = 2³ × 3³ × 2² × 5² = 2⁵ × 3³ × 5²</p><p><strong>Step 2:</strong> Distribute prime factors among x, y, z.</p><p>We need: x = 2^(a₁) × 3^(b₁) × 5^(c₁), y = 2^(a₂) × 3^(b₂) × 5^(c₂), z = 2^(a₃) × 3^(b₃) × 5^(c₃)</p><p>where a₁ + a₂ + a₃ = 5, b₁ + b₂ + b₃ = 3, c₁ + c₂ + c₃ = 2, and all exponents are non-negative integers.</p><p><strong>Step 3:</strong> Apply stars-and-bars for each prime independently.</p><p>For prime 2: Number of solutions to a₁ + a₂ + a₃ = 5 is C(5+3-1, 3-1) = C(7,2) = 21</p><p>For prime 3: Number of solutions to b₁ + b₂ + b₃ = 3 is C(3+3-1, 3-1) = C(5,2) = 10</p><p>For prime 5: Number of solutions to c₁ + c₂ + c₃ = 2 is C(2+3-1, 3-1) = C(4,2) = 6</p><p><strong>Step 4:</strong> Multiply the results (by fundamental counting principle).</p><p>Total solutions = 21 × 10 × 6 = 1260</p><p>∴ Answer: <strong>1260</strong></p>
Correct Answer: 1260