Question

Q28) If d is the HCF of 592 and 252 find x, y satisfying d = 592x + 252y. Also show that x and y are not unique integers.​

Answer 1

Answer:

4=47*252 - 20*592

Step-by-step explanation:

$592=2^4*37$

$252=2^2*3^2*7$

HCF=2^2=4

By division, we get

592 = 2*252 + 88

252 = 2*88 + 76

88 = 1*76 + 12

76 = 6 * 12 + 4

These equations can be writtemn as

4= 76 - 6*12........(1)

12= 88 - 1*76......(2)

76= 252 - 2*88.....(3)

88=592-2*252.......(4)

Now,

4= 76 - 6*12

4= 76 - 6*(88-1*76) (using (2))

4= 76 - 6*88 + 6*76

4=7*76 - 6*88

4=7*(252-2*88) - 6*88 (using (3))

4=7*252-14*88 - 6*88

4=7*252 - 20*88

4=7*252 - 20*(592-2*252) (using (4))

4=7*252 - 20*592+40*252

4=47*252 - 20*592 which is of the form 592x+252y