Question

If d id the hcf of 56 and 72, find x, y satisfyin d=56x+72y. also show that x and y are not unique

Answer 1

Using Euclid's division Algorithm

a = bq+ r

72= 56*1 + 16. -------(1)
56 = 16 * 3 + 8. -------(2)
16 = 8*2 + 0. -------(3)

Therefore 8 is the HCF
=> 8 = d

8 can be written as 56 - 16*3 { from 3}
=> 8 = 56 - 16*3
=> 8 = 56 - (72 -56*1) *3. { 16 = 72 - 56*1 from 1}
=> 8 = 56 - 72*3 +56 *3
=> 8 = 56 (1+3) +72(-3)
=> 8 = 56 (4) + 72 (-3)

=> 8 = 56x + 72y where x = 4 and y = -3

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