Question

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If d is the H.C.F of 56 and 72 , find x and y satisfying d = 56x+72y. Also , show that x and y are not unique.

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Answer 1

BY APPLYING EUCLID's DIVISION :

72 = 56 × 1 +16 -- (1)

56 = 16 × 3 + 8 -- (2)

16 = 8 × 2 + 0 -- (3)

THEREFORE HCF OF 56 AND 72 = 8

FROM (2) -- 8 = 56 - 16 ×3

8 = 56 - (72-56×1) ×3 (from 1)

8 = 56 - 3 × 72 + 56 × 3

8 = 56 × 4 + (-3) × 72

therefore , x = 4 and y = -3

Now, 8 = 56 × 4 + (-3) × 72

8 = 56 × 4 + (-3) × 72 - 56 × 72 + 56 × 72

8 = 56 × 4 -56 × 72 + (-3) × 72 + 56 ×72

8 = 56 × (4-72) + {(-3) + 56} × 72

8 = 56 × (-68) + (53) × 72

therefore x = -68 and y = 53

Answer 2

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