Question

d is the HCF of 65 and 117 expressed on the form of 65m + 117n.

Answer 1

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 !

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If you feel any difficulty you can use this also


Since 117 > 65, we apply the division lemma to 117 and 65 to obtain
117 = 65 � 1 + 52 … Step 1

Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain
65 = 52 � 1 + 13 … Step 2

Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain
52 = 4 � 13 + 0 … Step 3

In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers

The H.C.F. of 65 and 117 is 13

From Step 2:
13 = 65 – 52� 1 … Step 4

From Step 1:
52 = 117 – 65 � 1

Thus, from Step 4, it is obtained
13 = 65 – (117 – 65 � 1)
⇒13 = 65 � 2 – 117
⇒13 = 65 � 2 + 117 � (–1)

In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1