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If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y.
(Ans:5, -2 (Not unique)
Ans: Using Euclid's algorithm, the HCF (30, 72)
72 = 30 × 2 + 12
30 = 12 × 2 + 6
12 = 6 × 2 + 0
HCF (30,72) = 6
6=30-12×2
6=30-(72-30×2)2
6=30-2×72+30×4
6=30×5+72×-2
∴ x = 5, y = -2
Also 6 = 30 × 5 + 72 (-2) + 30 × 72 - 30 × 72
Solve it, to get
x = 77, y = -32
Hence, x and y are not unique
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If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y. (Ans:5, -2 (Not unique) Ans: Using Euclid's algorithm, the HCF (30, 72) 72 = 30 × 2 + 12
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