Question

HCF of 65 and 117 and find a pair of integral values of m and n such that HCF is 65m +117n​

Answer 1

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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Step-by-step explanation:

Answer 2

Answer: m=2 ; n=-1

Step by step explanation:

By Euclid's division lemma,

117=65×1+52

65=52x1+13

52=13x5+0

Hence, the HCF of 65 and 117 is 65 and 117.

Now going backwards,

13=65+52×(−1)

13=65+52×(−1)13=65+[117−65×1]×(−1)

13=65+52×(−1)13=65+[117−65×1]×(−1) [From (1)]

13=65+52×(−1)13=65+[117−65×1]×(−1) [From (1)]13=65×2+117×(−1)

13=65+52×(−1)13=65+[117−65×1]×(−1) [From (1)]13=65×2+117×(−1)∴m=2n=−1