Find HCF of 65 and 117 and express it in the form 65m+117n
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Between 65 and 117; 117 is greater than 65
Division lemma of 117 and 65:
⇒ 117 = 65 x 1 + 52
⇒ 65 = 52 x 1 + 13
⇒ 52 = 4 x 13 + 0
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
Now, 13 = 65 – 52 x 1
⇒ 52 = 117 – 65 x 1
⇒ 13 = 65 – (117 – 65 x 1) x 1
⇒ 13 = 65 x 2 – 117
⇒ 13 = 65 x 2 + 117 x (–1)
Therefore H.C.F. of 65 and 117 is of the form 65m + 117n, where m = 2 and n = –1.
Division lemma of 117 and 65:
⇒ 117 = 65 x 1 + 52
⇒ 65 = 52 x 1 + 13
⇒ 52 = 4 x 13 + 0
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
Now, 13 = 65 – 52 x 1
⇒ 52 = 117 – 65 x 1
⇒ 13 = 65 – (117 – 65 x 1) x 1
⇒ 13 = 65 x 2 – 117
⇒ 13 = 65 x 2 + 117 x (–1)
Therefore H.C.F. of 65 and 117 is of the form 65m + 117n, where m = 2 and n = –1.
GeniusShweta:
correct
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