If d is the HCF of 56 and 72, find x, y satisfying d = 56x + 72y. Also, show that x and y are not unique.
Answers
Let d be the HCF of 56 and 72
Let a = 72 and b = 56
By Euclid's division Lemma:
a = bq + r
=> 72 = 56 × 1 + 16 ----> (1)
=> 56 = 16 × 3 + 8 ----> (2)
=> 16 = 8 × 2 + 0 ----> (3)
∴ HCF(56, 72) = 8
we can also write,
d = 8
=> d = 56 - 16 × 3 {from (2)}
=> d = 56 - (72 - 56 × 1) × 3 {from (1)}
=> d = 56 - (72 - 56) × 3
We are not going to subtract the part in the bracket, Since we have to express them as 56x and 72y..So just multiply by 3
=> d = 56 - 72 × 3 + 56 × 3
=> d = 56 + 56 × 3 - 72 × 3
Taking 56 as common
=> d = 56(1 + 3) - 72 × 3
=> d = 56 × 4 - 72 × 3
We can also write -72 × 3 as 72 × (-3).. Since the product will be same
=> d = 56 × 4 + 72 × (-3)
Comparing with the form d = 56x + 72y
We get,
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
HENCE , x and y are not unique
Step-by-step explanation:
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