<p>Find three-digit numbers that are divisible by 5 as well as 9 and whose consecutive digits are in AP.</p>
Step-by-Step Solution
Key Concept: A three-digit number with consecutive digits in AP can be expressed as (a-d), a, (a+d), and must satisfy divisibility conditions for both 5 and 9. The divisibility by 5 restricts the units digit, while divisibility by 9 requires the sum of digits to be divisible by 9.
<p><strong>Step 1: Set up the three-digit number with digits in AP.</strong></p><p>Let the three digits be (a-d), a, (a+d) where a is the middle digit and d is the common difference. The number is: 100(a-d) + 10a + (a+d) = 111a - 99d.</p><p><strong>Step 2: Apply divisibility by 5 condition.</strong></p><p>For divisibility by 5, the units digit must be 0 or 5. Therefore: (a+d) = 0 or (a+d) = 5.</p><p><strong>Case 1: a + d = 0</strong></p><p>Then d = -a, and the digits are (2a), a, 0. For valid digits: 0 ≤ 2a ≤ 9, so a ∈ {1,2,3,4}.</p><p><strong>Step 3: Apply divisibility by 9 condition for Case 1.</strong></p><p>Sum of digits: 2a + a + 0 = 3a must be divisible by 9. So 3a ≡ 0 (mod 9), giving a ≡ 0 (mod 3). Thus a ∈ {3}.</p><p>When a = 3: digits are 6, 3, 0 → number is 630. Check: 630 ÷ 5 = 126 ✓, 630 ÷ 9 = 70 ✓</p><p><strong>Case 2: a + d = 5</strong></p><p>Then d = 5 - a, and the digits are (2a-5), a, 5. For valid digits: 0 ≤ 2a-5 ≤ 9 and 1 ≤ a ≤ 9, so a ∈ {3,4,5,6,7}.</p><p><strong>Step 4: Apply divisibility by 9 condition for Case 2.</strong></p><p>Sum of digits: (2a-5) + a + 5 = 3a must be divisible by 9. So a ≡ 0 (mod 3), giving a ∈ {3, 6}.</p><p>When a = 3: d = 2, digits are 1, 3, 5 → number is 135. Check: 135 ÷ 5 = 27 ✓, 135 ÷ 9 = 15 ✓</p><p>When a = 6: d = -1, digits are 7, 6, 5 → number is 765. Check: 765 ÷ 5 = 153 ✓, 765 ÷ 9 = 85 ✓</p><p><strong>Step 5: Verify all constraints.</strong></p><p>All three numbers satisfy: three-digit format, divisible by 5, divisible by 9, and consecutive digits in AP.</p><p><strong>∴ Answer: 135, 630, 765</strong></p>
Correct Answer: 135