Question

If the hcf of 152 and 272 is expressible in the form 272y + 152x, then find x and y

Answer 1

152 = 2*2*2*19
272 =2*2*2*2*17
so hcf is 2*2*2=8
now 8=272y+152x
since a= bq+r
or 272=152*1+(272-152)=152+120
now
272*8+152x=hcf
or 2176+152x=8
or x= -2168/152= -271/19
so 8= 272*8+152*(-271/19)
so x=(-271/19) and y=8

Answer 2

Answer:

Step-by-step explanation:

⇒ 272 = (152*1) + 120

⇒ 152 = (120*1) + 32

⇒ 120 = (32*3) + 24

⇒ 32 = (24*1) + 8

⇒ 24 = (8*3) + 0

Since the remainder becomes 0 here, so the H.C.F. of 152 and 272 is 8

Now,

272*8 + 152x = H.C.F. of these numbers

⇒ 2176 + 152x = 8

⇒ 152x = 8 - 2176

⇒ 152x = - 2168

⇒ x = - 2168/152

⇒ x = - 271/19