<p>9000! is divisible by \(10^n\). Find the largest positive integral value of \(n\).</p>
Step-by-Step Solution
Key Concept: The number of trailing zeros in n! equals the highest power of 10 dividing n!, which is determined by the highest power of 5 dividing n! (since factors of 2 are always more abundant). Use Legendre's formula: count ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...
<p><strong>Step 1:</strong> To find the highest power of 10 dividing 9000!, we need the highest power of 5 (since factors of 2 are always more abundant).</p><p><strong>Step 2:</strong> Apply Legendre's formula for prime p = 5:</p><p>n = ⌊9000/5⌋ + ⌊9000/25⌋ + ⌊9000/125⌋ + ⌊9000/625⌋ + ⌊9000/3125⌋ + ⌊9000/15625⌋</p><p><strong>Step 3:</strong> Calculate each term:</p><p>• ⌊9000/5⌋ = 1800</p><p>• ⌊9000/25⌋ = 360</p><p>• ⌊9000/125⌋ = 72</p><p>• ⌊9000/625⌋ = 14</p><p>• ⌊9000/3125⌋ = 2</p><p>• ⌊9000/15625⌋ = 0 (stop here)</p><p><strong>Step 4:</strong> Sum all contributions:</p><p>n = 1800 + 360 + 72 + 14 + 2 = 2248</p><p>∴ Answer: <strong>2248</strong></p>
Correct Answer: 2248