Question

If the hcf of 152 and 272 is expressible in the form 272y + 152x, find the value of x and y also show that x and y are not unique.

Answer 1

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Answer 2

Solution :-

H.C.F. of 152 and 272 by Euclid's Division Algorithm.

⇒ 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

Answer.
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