<p>An \(n\)-digit number is a positive number with exactly \(n\) digits. Nine hundred distinct \(n\)-digit numbers are to be formed using only the three digits 2, 5 and 7. Find the smallest value of \(n\) for which this is possible.</p>
Step-by-Step Solution
Key Concept: The maximum count of n-digit numbers using only 3 digits is 3^n (each position has 3 choices). We need 3^n ≥ 900, so find the smallest n satisfying this inequality.
<p><strong>Step 1:</strong> Identify the constraint. We have 3 available digits {2, 5, 7} and each position in an n-digit number can be filled with any of these 3 digits.</p><p><strong>Step 2:</strong> Calculate total n-digit numbers possible. For an n-digit number, we have n positions, and each can be filled in 3 ways. Total = 3^n.</p><p><strong>Step 3:</strong> Set up inequality. We need at least 900 distinct numbers: 3^n ≥ 900.</p><p><strong>Step 4:</strong> Calculate powers of 3:<br/>3^1 = 3<br/>3^2 = 9<br/>3^3 = 27<br/>3^4 = 81<br/>3^5 = 243<br/>3^6 = 729 < 900<br/>3^7 = 2187 ≥ 900 ✓</p><p><strong>Step 5:</strong> Verify. 3^6 = 729 < 900 (insufficient), but 3^7 = 2187 ≥ 900 (sufficient).</p><p>∴ Answer: <strong>7</strong></p>
Correct Answer: 7