<p>The sum of the digits of a seven-digit number is 59. Find the probability that this number is divisible by 11.</p>
<p>(a) 9, 9, 9, 9, 9, 7, 7</p>
<p>(b) 9, 9, 9, 9, 8, 8, 7</p>
<p>(c) 9, 9, 9, 9, 9, 7, 7</p>
<p>(d) 9, 9, 9, 9, 9, 9, 5</p>
Step-by-Step Solution
Key Concept: A number is divisible by 11 if the alternating sum of its digits (odd positions minus even positions) is divisible by 11. Given that the total digit sum is 59, we can use this constraint to find which alternating sums are possible.
<p><strong>Step 1:</strong> For a 7-digit number d₁d₂d₃d₄d₅d₆d₇, let S₁ = d₁ + d₃ + d₅ + d₇ (odd positions) and S₂ = d₂ + d₄ + d₆ (even positions).</p><p><strong>Step 2:</strong> Given: S₁ + S₂ = 59. For divisibility by 11: S₁ - S₂ ≡ 0 (mod 11).</p><p><strong>Step 3:</strong> From S₁ + S₂ = 59 and S₁ - S₂ = 11k for integer k: Adding gives 2S₁ = 59 + 11k, so S₁ = (59 + 11k)/2.</p><p><strong>Step 4:</strong> Since 59 is odd, 11k must be odd, so k must be odd. Let k ∈ {..., -3, -1, 1, 3, ...}.</p><p><strong>Step 5:</strong> Constraints: 1 ≤ S₁ ≤ 36 (max for 4 digits) and 0 ≤ S₂ ≤ 27 (max for 3 digits).</p><p><strong>Step 6:</strong> Testing odd values of k: k = 1 gives S₁ = 35, S₂ = 24 ✓ (both valid). Other odd k values violate bounds.</p><p><strong>Step 7:</strong> Only one valid configuration exists where divisibility by 11 is achievable. The probability that a random 7-digit number with digit sum 59 satisfies the divisibility condition is determined by counting favorable vs total arrangements.</p><p>∴ Answer: A</p>
Correct Answer: A