If the hcf of 152 and 272 is expressible in the form 272 x 8 + 152x, then find x
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168
Here we find the HCF by Euclid Division Lemma (a=bq+r)
The value of x is (-2168/152)
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Hope this will help you...
The value of x is (-2168/152)
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Hope this will help you...
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Answered by
274
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.
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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