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How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 1080?
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We need to find ordered pairs (x, y) such that xy * HCF(x, y) = 1080. Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1. So h3(ab) = 1080 = (23)(33)(5). We need to write 1080 as a product of a perfect cube and another number. Four cases: 1. h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1, 1080), (8, 135), (27, 40) and (5, 216). (Essentially we are finding co-prime a,b such that a*b = 1080). 2. h = 2, We need to find number of ways of writing (33) * (5) as a product of two co-prime numbers. This can be done in two ways - 1 and (33) * (5) , (33) and (5) number of pairs = 2, number of ordered pairs = 4 3. h = 3, number of pairs = 2, number of ordered pairs = 4 4. h = 6, number of pairs = 1, number of ordered pairs = 2 Hence total pairs of (x, y) = 9, total number of ordered pairs = 18. The pairs are (1, 1080), (8, 135), (27, 40), (5, 216), (2, 270), (10, 54), (3, 120), (24, 15) and (6, 30).
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