The correct option is B 2
Let the HCF of 65 and 117 be x.
Here, 65 = 13 × 5
117 = 13 × 3 × 3
⇒ LCM of 65 and 117 = 585
Now, product of 65 and 117 = HCF(65, 117) × LCM(65, 117)
⇒ HCF(65, 117) × 585 = 65 × 117
⇒ HCF(65, 117) = 1179
⇒ HCF(65, 117) = 13
Now, HCF(65, 117) = 13 = 65m - 117
⇒ m = 13065=2
Alternate Solution
By prime factorisation ;
117 = 3 × 3 × 13
65 = 5 × 13
Hence HCF of 117 and 65 is 13
HCF(65, 117) = 13 = 65m - 117
Hence m = 2