<p>How many three – term harmonic progressions a, b, c of strictly increasing positive integers in which \(a = 20\) and \(b\) divides \(c\) exist?</p>
Step-by-Step Solution
Key Concept: A harmonic progression (HP) is the reciprocal of an arithmetic progression. For three terms a, b, c in HP, we have 1/a, 1/b, 1/c in AP, which gives 2/b = 1/a + 1/c. This relationship must be combined with the constraints: a = 20, a < b < c (strictly increasing positive integers), and b | c.
<p><strong>Step 1:</strong> Express the harmonic progression condition. If a, b, c are in HP, then 1/a, 1/b, 1/c are in AP, so:</p><p>2/b = 1/a + 1/c</p><p><strong>Step 2:</strong> Substitute a = 20:</p><p>2/b = 1/20 + 1/c</p><p>2/b = (c + 20)/(20c)</p><p>40c = b(c + 20)</p><p>40c = bc + 20b</p><p>40c - bc = 20b</p><p>c(40 - b) = 20b</p><p>c = 20b/(40 - b)</p><p><strong>Step 3:</strong> For c to be a positive integer, (40 - b) must divide 20b. Let d = 40 - b, so b = 40 - d. Then:</p><p>c = 20(40 - d)/d = (800 - 20d)/d = 800/d - 20</p><p>For c to be a positive integer, d must divide 800.</p><p><strong>Step 4:</strong> Apply constraints. We need: 20 < b < c and b | c.</p><p>From b = 40 - d: 20 < 40 - d gives d < 20</p><p>From b < c: 40 - d < 800/d - 20, so 60 - d < 800/d, giving d(60 - d) < 800, or 60d - d² < 800</p><p><strong>Step 5:</strong> Find divisors of 800 = 2⁵ × 5² that are less than 20: {1, 2, 4, 5, 8, 10, 16}</p><p><strong>Step 6:</strong> For each divisor d, calculate b = 40 - d and c = 800/d - 20, then verify b < c and b | c:</p><p>• d = 1: b = 39, c = 780. Check: 39 < 780 ✓, 780/39 = 20 ✓</p><p>• d = 2: b = 38, c = 380. Check: 38 < 380 ✓, 380/38 = 10 ✓</p><p>• d = 4: b = 36, c = 180. Check: 36 < 180 ✓, 180/36 = 5 ✓</p><p>• d = 5: b = 35, c = 140. Check: 35 < 140 ✓, 140/35 = 4 ✓</p><p>• d = 8: b = 32, c = 80. Check: 32 < 80 ✓, 80/32 = 2.5 ✗</p><p>• d = 10: b = 30, c = 60. Check: 30 < 60 ✓, 60/30 = 2 ✓</p><p>• d = 16: b = 24, c = 30. Check: 24 < 30 ✓, 30/24 = 1.25 ✗</p><p><strong>∴ Answer: 5</strong></p>
Correct Answer: 5