Question

find hcf of 65 and 117 and find pair of integral values of m and n such that hcf=65m and 117n​

Answer 1

Answer:

Applying Euclid's division algorithm,

117=65×1+52 ---(1)

65=52×1+13 ---(2)

52=13×4+0

Therefore, HCF(65,117) = 13

Now,

13=65m+117n

from values of (1) and (2),

13=65-52×1

52=117-65×1 ---(3)

now,

13=65-(117-65×1)

13=65-117+65

13=65(2) - 117(1)

13=65(2)+117(-1)

comparing with 65m and 117n,

m = 2

n = -1

Although this question is most probably deleted as it includes Euclid's division algorithm, but still

hope it helps