HCF of 152 and 272 is expressible in the form of 272 into 8 + 152 P then find p
ManushkasinghM:
is it 272(8 + 152p)
Answers
Answered by
2
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
⇒ 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
Similar questions