Find the HCF of 65 and 117 and express it in form of 65m +117n
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Answered by
2173
Hello there !
Euclid's Division Lemma :-
a = bq +r
117 > 65
117 = 65 × 1 + 52 ----> [ 2 ]
65 = 52 x 1 + 13 -----> [1]
52 = 13 x 4 + 0
HCF = 13
13 = 65m + 117n
From [ 1] ,
13 = 65 - 52 x 1
From [2] ,
52 = 117 - 65 x 1 ----> [3]
Hence ,
13 = 65 - [ 117 - 65 x 1 ] ------> from [3]
= 65 x 2 - 117
= 65 x 2 + 117 x [-1 ]
m = 2
n = -1
Hope this Helped You !
Euclid's Division Lemma :-
a = bq +r
117 > 65
117 = 65 × 1 + 52 ----> [ 2 ]
65 = 52 x 1 + 13 -----> [1]
52 = 13 x 4 + 0
HCF = 13
13 = 65m + 117n
From [ 1] ,
13 = 65 - 52 x 1
From [2] ,
52 = 117 - 65 x 1 ----> [3]
Hence ,
13 = 65 - [ 117 - 65 x 1 ] ------> from [3]
= 65 x 2 - 117
= 65 x 2 + 117 x [-1 ]
m = 2
n = -1
Hope this Helped You !
Answered by
558
Answer:
By Euclid's division algorithm
117 = 65x1 + 52.
65 = 52x1 + 13
52 = 13x4 + 0
Therefore 13 is the HCF (65, 117).
Now work backwards:
13 = 65 + 52x(-1)
13 = 65 + [117 + 65x(-1)]x(-1)
13 = 65x(2) + 117x(-1).
∴ m = 2 and n = -1.
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