<p>Consider a 101 digit number \(N = pa_1a_2a_3\ldots a_{99}a_{100}\) where \(p\) is a prime digit and \(a_i\) is any digit. How many numbers \(N\) can be formed such that digit 9 is used odd number of times is \(k\), then \(k\) is</p>
<p>an even number</p>
<p>a perfect square number</p>
<p>is divisible by \(2^{16}\)</p>
<p>not divisible by \(2^{32}\)</p>
Step-by-Step Solution
Key Concept: Use the complementary counting principle: total numbers with 9 appearing odd times = (total numbers) - (numbers with 9 appearing even times). For each position, track parity of 9's count using the generating function approach or binary splitting of digits into {9} and {0-8,10...}.
<p><strong>Step 1:</strong> Identify the structure. We have 101 digits total: first digit p (prime: 2,3,5,7) and 100 remaining digits (a₁ to a₁₀₀), each from 0-9.</p><p><strong>Step 2:</strong> Count total numbers: p has 4 choices, each aᵢ has 10 choices. Total = 4 × 10¹⁰⁰</p><p><strong>Step 3:</strong> Use complementary counting. For any set of positions, numbers with 9 appearing even times = numbers with 9 appearing odd times (when total is evenly split). This works because for 100 free digits (a₁ to a₁₀₀), exactly half of all 10¹⁰⁰ arrangements have 9 appearing an odd number of times.</p><p><strong>Step 4:</strong> For the 100 positions {a₁,...,a₁₀₀}: Among 10¹⁰⁰ total arrangements, by parity principle, exactly 10¹⁰⁰/2 have odd count of 9's and 10¹⁰⁰/2 have even count of 9's.</p><p><strong>Step 5:</strong> The first digit p ∈ {2,3,5,7} does NOT contain 9, so it doesn't affect the parity of 9's count.</p><p><strong>Step 6:</strong> Therefore, k = 4 × (10¹⁰⁰/2) = 4 × 5 × 10⁹⁹ = 2 × 10¹⁰⁰</p><p>∴ Answer: A</p>
Correct Answer: A