<p>How many natural numbers are there lying between 20,000 and 60,000, the sum of digits being even?</p>
Step-by-Step Solution
Key Concept: A number has even digit sum if and only if it contains an even number of odd digits. For 5-digit numbers in the range [20000, 60000), partition by first digit (2, 3, 4, 5) and count arrangements where the remaining 4 digits sum to even parity.
<p><strong>Step 1: Set up the range</strong></p><p>Numbers between 20,000 and 60,000 have the form 2d₂d₃d₄d₅, 3d₂d₃d₄d₅, 4d₂d₃d₄d₅, or 5d₂d₃d₄d₅ where each dᵢ ∈ {0,1,...,9}.</p><p><strong>Step 2: Determine parity constraint</strong></p><p>For even digit sum: if first digit is even (2 or 4), remaining 4 digits must sum to even; if first digit is odd (3 or 5), remaining 4 digits must sum to odd.</p><p><strong>Step 3: Count for even first digit (2 or 4)</strong></p><p>For 4 digits d₂d₃d₄d₅, we need an even number of odd digits among them.</p><p>Total arrangements: 10⁴ = 10,000</p><p>By symmetry, exactly half have even parity and half have odd parity: 5,000 each.</p><p>For first digit 2: 5,000 numbers</p><p>For first digit 4: 5,000 numbers</p><p><strong>Step 4: Count for odd first digit (3 or 5)</strong></p><p>For 4 digits d₂d₃d₄d₅, we need an odd number of odd digits.</p><p>By symmetry: 5,000 numbers each</p><p>For first digit 3: 5,000 numbers</p><p>For first digit 5: 5,000 numbers</p><p><strong>Step 5: Total count</strong></p><p>5,000 + 5,000 + 5,000 + 5,000 = 20,000</p><p>However, we must exclude 60,000 if included. Since 60,000 is not in (20,000, 60,000), and our range counts [20,000, 60,000), we subtract 1 (the number 20,000 itself has digit sum 2, which is even, so it's counted).</p><p>Rechecking: numbers in [20,000, 60,000) with even digit sum = 4 × 5,000 - 1 = <strong>19,999</strong></p><p>∴ Answer: 19999</p>
Correct Answer: 19999