Question

If d is the hcf of 56 and 72 find x and y satisfying d=56x +72y.

Answer 1

Euclid division lemma:-

a = bq + r

0 ≤ r < b

a > b

72 > 56

72 = 56 × 1 + 16

56 = 16 × 3 + 8

16 = 8 × 2 + 0

As the remainder is 0,HCF is 8

HCF of 56 and 72 is 8

d = 8

d = 56x + 72y

8 = 56 - 16×3

= 56 - [72 - 56(1)]×3

= 56 - 72×3 + 56×3

= 56×4 - 72×3

= 56×4 + 72(-3)

= 56x + 72y

Therefore, x = 4 and y = -3

Hope it helps

Answer 2

Answer:-

$\begin{lgathered}\begin{lgathered}\begin{lgathered}72 = 56 \times 1 = 16 \\ 56 = 16 \times 3 + 8 \\ 16 = 8 \times 2 + 10\end {lgathered}\end{lgathered}\end{lgathered}$

Now L. C. M =8

72−56=16

56−(72−56)×3=8

8=56+72(−3)+56(3)

8=56(4)+72(−3)

d=56x+72y

x=4andy=−3

(72)(56)−(72)(56)

8=56(4)+72(−3)+(72)(56)−(72)(56)

8=56(4+72)+72(−3−56)

8=56(76)+72(−59)

d=56x+72y

Therefore, x=56 andy = 59 x and are not unique