<p>How many 10-digit numbers can be made with odd digits so that no two consecutive digits are the same?</p>
Step-by-Step Solution
Key Concept: Use recursion/counting principle: if a(n) is the count of valid n-digit numbers, establish that a(n) = 4·a(n-1) because each valid (n-1)-digit number can be extended with 4 different odd digits (all except the last digit). Start with a(1) = 5 base case.
<p><strong>Step 1:</strong> Identify constraints. We need 10-digit numbers using only odd digits {1, 3, 5, 7, 9} where no two consecutive digits are identical.</p><p><strong>Step 2:</strong> Set up recurrence. Let a(n) = number of valid n-digit numbers.</p><p><strong>Step 3:</strong> For the first digit: 5 choices (any odd digit).</p><p><strong>Step 4:</strong> For each subsequent digit: 4 choices (any odd digit except the previous one).</p><p><strong>Step 5:</strong> The recurrence relation is: a(n) = 5 × 4^(n-1)</p><p><strong>Step 6:</strong> For n = 10: a(10) = 5 × 4^9</p><p><strong>Step 7:</strong> Calculate 4^9 = 262,144</p><p><strong>Step 8:</strong> Therefore: a(10) = 5 × 262,144 = <strong>1,310,720</strong></p><p>∴ Answer: <strong>5 × 4^9 = 1,310,720</strong></p>
Correct Answer: 5