<p>If \((1 + 3 + 5 + \cdots + p) + (1 + 3 + 5 + \cdots + q) = (1 + 3 + 5 + \cdots + r)\) where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of \(p + q + r\) (where \(p > 6\)) is</p>
Step-by-Step Solution
Key Concept: The sum of the first n odd integers equals n². Convert each sum to the form n², then solve n²_p + n²_q = n²_r where n_p, n_q, n_r are positive integers—this is finding Pythagorean triples.
<p><strong>Step 1:</strong> Recognize that the sum of first n consecutive odd integers = n².</p><p>If 1+3+5+...+p contains m terms, then p = 2m-1 and the sum = m².</p><p><strong>Step 2:</strong> Let the three sums contain a, b, c terms respectively. Then a² + b² = c² (Pythagorean triple).</p><p><strong>Step 3:</strong> Since p > 6, we need p = 2a-1 > 6, so a ≥ 4. The smallest Pythagorean triple with a ≥ 4 is (3,4,5).</p><p><strong>Step 4:</strong> For a=4: p = 2(4)-1 = 7 ✓</p><p>For (3,4,5) triple: b=3, so q = 2(3)-1 = 5; and c=5, so r = 2(5)-1 = 9.</p><p><strong>Step 5:</strong> Verify: (1+3+5+7) + (1+3+5) = 16 + 9 = 25 = (1+3+5+7+9) ✓</p><p>Therefore p + q + r = 7 + 5 + 9 = <strong>21</strong></p><p>∴ Answer: D</p>
Correct Answer: D