<p><strong>For Problems 1–3</strong><br>Sum of certain consecutive odd positive integers is \(57^2 - 13^2\).</p><p><strong>Problem 1:</strong> Number of integers are</p>
Step-by-Step Solution
Key Concept: Factor the difference of squares as (57-13)(57+13) = 44×70, then recognize this equals the sum of n consecutive odd integers starting from (2a+1), which gives n² = 3080 or use the formula: sum of n consecutive odd integers = n². However, 3080 isn't a perfect square, so we must find which n divides 3080 such that consecutive odd integers sum to exactly 3080.
<p><strong>Step 1:</strong> Calculate 57² - 13² using difference of squares.</p><p>57² - 13² = (57-13)(57+13) = 44 × 70 = 3080</p><p><strong>Step 2:</strong> Sum of n consecutive odd integers starting from (2a+1) is:</p><p>S = n(2a+n) = 3080, where a ≥ 0</p><p><strong>Step 3:</strong> Factor 3080 = 8 × 385 = 8 × 5 × 77 = 40 × 77</p><p>For n and (2a+n) to both be positive integers with a ≥ 0:</p><p>• If n = 40: then 2a + 40 = 77, so a = 18.5 ✗ (not integer)</p><p>• If n = 44: then 2a + 44 = 70, so a = 13 ✓</p><p>• If n = 70: then 2a + 70 = 44, so a = -13 ✗ (a must be ≥ 0)</p><p>• If n = 77: then 2a + 77 = 40, so a < 0 ✗</p><p><strong>Step 4:</strong> Verify: 44 consecutive odd integers starting from (2×13+1) = 27</p><p>Sum = 27 + 29 + 31 + ... + 113 = 44(27+113)/2 = 44 × 70 = 3080 ✓</p><p>∴ Answer: <strong>44</strong> (Option C)</p>
Correct Answer: C