<p>Find the sum of all the numbers greater than 10,000 that can be made with the digits 1, 3, 5, 7 and 9 if digits are not repeated in the same number.</p>
Step-by-Step Solution
Key Concept: Numbers greater than 10,000 must have at least 5 digits. Calculate the sum by finding how many times each digit appears in each position across all valid permutations, then sum contributions by position value.
<p><strong>Step 1:</strong> Identify valid numbers. With digits {1, 3, 5, 7, 9}, numbers greater than 10,000 must be exactly 5-digit numbers using all digits without repetition. Total such numbers = 5! = 120.</p><p><strong>Step 2:</strong> Use symmetry principle. Each of the 5 digits appears equally often in each position (units, tens, hundreds, thousands, ten-thousands). In each position, each digit appears 120/5 = 24 times.</p><p><strong>Step 3:</strong> Calculate sum of digits = 1 + 3 + 5 + 7 + 9 = 25.</p><p><strong>Step 4:</strong> Sum contribution by position:<br/>• Ten-thousands place: 24 × 25 × 10,000 = 6,000,000<br/>• Thousands place: 24 × 25 × 1,000 = 600,000<br/>• Hundreds place: 24 × 25 × 100 = 60,000<br/>• Tens place: 24 × 25 × 10 = 6,000<br/>• Units place: 24 × 25 × 1 = 600</p><p><strong>Step 5:</strong> Total sum = 6,000,000 + 600,000 + 60,000 + 6,000 + 600 = 6,666,600.</p><p><strong>Alternatively:</strong> Sum = 24 × 25 × (10,000 + 1,000 + 100 + 10 + 1) = 24 × 25 × 11,111 = 600 × 11,111 = 6,666,600</p><p>∴ <strong>Answer: 6,666,600</strong></p>
Correct Answer: 6