Question

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

Answer 1

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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

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

2176 + 152x = 8

152x = 8 - 2176

152x = - 2168

x = - 2168/152

x = - 271/19