<p>Find the number of six-digit numbers that can be formed using the digits 0, 1, 2, 5, 7, 9 such that the number is divisible by 11.</p>
Step-by-Step Solution
Key Concept: Apply divisibility rule for 11 (difference of alternately placed digit sums) and partition the given digits into two sets with equal sum.
<p><strong>Step 1:</strong> Sum of digits = 0 + 1 + 2 + 5 + 7 + 9 = 24</p><p><strong>Step 2:</strong> For divisibility by 11, |(a + c + e) - (b + d + f)| must be 0 or a multiple of 11.</p><p><strong>Step 3:</strong> Only possible case is a + c + e = 12 = b + d + f</p><p><strong>Step 4:</strong> Case I: {a, c, e} = {0, 5, 7} and {b, d, f} = {1, 2, 9}<br/>Number of 6-digit numbers = (2 × 2!) × 3! = 24 (since a cannot be 0)</p><p><strong>Step 5:</strong> Case II: {a, c, e} = {1, 2, 9} and {b, d, f} = {0, 5, 7}<br/>Number of 6-digit numbers = 3! × 3! = 36</p><p><strong>Step 6:</strong> Total = 24 + 36 = 60</p>
Correct Answer: 60