<p>Five different digits from the set of numbers \(\{1, 2, 3, 4, 5, 6, 7\}\) are written in random order. How many numbers can be formed using 5 different digits from this set if the number is divisible by 9?</p>
Step-by-Step Solution
Key Concept: A number is divisible by 9 if and only if the sum of its digits is divisible by 9. First identify which 5-digit subsets from {1,2,3,4,5,6,7} have digit sum divisible by 9, then count permutations of each valid subset.
<p><strong>Step 1:</strong> Find sum of all digits: 1+2+3+4+5+6+7 = 28</p><p><strong>Step 2:</strong> For 5 digits selected, we exclude 2 digits. If we exclude digits with sum S, then digit sum = 28 - S. We need 28 - S ≡ 0 (mod 9), so S ≡ 28 ≡ 1 (mod 9).</p><p><strong>Step 3:</strong> Find all pairs of excluded digits whose sum ≡ 1 (mod 9):</p><ul><li>1+9: not possible (9 not in set)</li><li>2+8: not possible (8 not in set)</li><li>3+7: sum = 10 ≡ 1 (mod 9) ✓ → remaining digits {1,2,4,5,6}, sum = 18</li><li>4+6: sum = 10 ≡ 1 (mod 9) ✓ → remaining digits {1,2,3,5,7}, sum = 18</li></ul><p><strong>Step 4:</strong> Verify: 18 ≡ 0 (mod 9) ✓</p><p><strong>Step 5:</strong> We have exactly 2 valid 5-digit subsets, each with digit sum = 18 (divisible by 9).</p><p><strong>Step 6:</strong> For each valid subset, the number of permutations = 5! = 120</p><p><strong>Step 7:</strong> Total numbers = 2 × 120 = 240</p><p>∴ Answer: <strong>240</strong></p>
Correct Answer: 240