<p>Find the sum of all natural numbers \(n\) such that \(1000 \leq n \leq 4000\), that can be made with the digits 0, 1, 2, 3, 4 if repetition of digits in the same number is allowed.</p>
Step-by-Step Solution
Key Concept: Numbers from 1000-4000 have fixed first digit (1, 2, or 3), and remaining positions can be filled freely with {0,1,2,3,4}. Use the principle that sum = (count of numbers) × (average value) by treating each position's contribution separately.
<p><strong>Step 1: Count valid numbers by first digit</strong></p><p>First digit ∈ {1, 2, 3} (3 choices)</p><p>Remaining three digits ∈ {0, 1, 2, 3, 4} each (5 choices each)</p><p>Total count = 3 × 5 × 5 × 5 = 3 × 125 = 375 numbers</p><p><strong>Step 2: Calculate sum by position contribution</strong></p><p><u>Thousands place:</u></p><p>Digit 1 appears: 5³ = 125 times</p><p>Digit 2 appears: 5³ = 125 times</p><p>Digit 3 appears: 5³ = 125 times</p><p>Contribution = 1000(125×1 + 125×2 + 125×3) = 1000 × 125 × 6 = 750,000</p><p><u>Hundreds place:</u></p><p>Each digit {0,1,2,3,4} appears equally: 3 × 5² = 75 times each</p><p>Sum of digits = 0+1+2+3+4 = 10</p><p>Contribution = 100 × 75 × 10 = 75,000</p><p><u>Tens place:</u></p><p>Each digit {0,1,2,3,4} appears: 3 × 5² = 75 times each</p><p>Contribution = 10 × 75 × 10 = 7,500</p><p><u>Units place:</u></p><p>Each digit {0,1,2,3,4} appears: 3 × 5² = 75 times each</p><p>Contribution = 1 × 75 × 10 = 750</p><p><strong>Step 3: Total sum</strong></p><p>Sum = 750,000 + 75,000 + 7,500 + 750 = 833,250</p><p>∴ Answer: <strong>833,250</strong></p>
Correct Answer: 833