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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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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
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