LCM of 6 and 10 would be the first number which would be divisible by both 6 and 10
LCM is 30.
All the multiples of 30 would also satisfy the problem.
Case 1: considering 30 as N then A would result in 0 as remainder of 30/30 is 0
For B, remainder of 30/60 would result in 30
Therefore, A<B
Case 2: Considering 60 (a multiple of 30) as N
A: 0 (remainder of 60/30=0)
B: 0 (remainder of 60/60=0)
Therefore A = B

Correct Option D.