Question

If the HCF of 152 and 272 is expressible in the form
272 X8+ 152x,then find x
​

Answer 1

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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Answer 2

Answer:

Step-by-step explanation:

by  euclids lemma

272 = 152 x 1 + 120

152 = 120 x 1 + 32

120 = 32 x 3 + 24

32 = 24 x 1 + 8

24 = 8 x3 + 0

thus H.C.F is 8

272 x 8 + 152 x X = 8

152(x) = 2168

x = 2168/152

x = 14