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If d is the HCF of $56$ and $72,$ then $d = 56 x + 72 y$ . Here $x y$ are not unique.

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True

False

answer is A.

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Detailed Solution

Given,

d = 56 x + 72 y

Euclid Division Lemma is,

a = bq + r

Here,

r =

Remainder

b =

Divisor

a =

Dividend

q =

Quotient
From Euclid Division Lemma,

72 = 56 \times 1 + 16

56 = 16 \times 3 + 8

16 = 8 \times 2 + 0

So, highest common factor of

56 72

8

.
So, d is 8. Thus,

d = 56 x + 72 y

56 x + 72 y = 8

…..(1)
Now, we have,

8 \times 2 = 16

Also,

(56 - 16 \times 3) \times 2 = 16 (56 - 16 \times 3) = 8

Similarly,

(56 - (72 - 56) \times 3) = 8

56 \times 4 - 72 \times 3 = 8

(56) (4) + (72) (- 3) = 8 . . . . . . (2)

On comparing

(1) (2),

we get

x = 4 y = - 3

Equation 2 can be rewritten as,

(56) (4) + (72) (- 3) + (56 \times 72) - (56 \times 72) = 8

56 \times 4 + 56 \times 72 + 72 \times (- 3) + 72 \times (- 56) = 8 56 \times (76) + 72 \times (- 59) = 8

Compare this equation with (1), we get

x = 76, y = - 59

One equation with two variable results in no fixed value of x and y. Thus,

x y

are not unique.
Option 1 is Correct.

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