So x can be written as 5p+1, where p is a non-negative Integer1). When x is divided by 5, the remainder is 1
If p = 0, x = 1 and is not divisible by 3
If p = 1, x =6 and is divisible by 3
Hence Insufficient!
So x can be written as 15p+1, where p is a non-negative Integer2). When x is divided by 15, the remainder is 1
15p+1 is one more than a multiple of 3. So, is not divisible by 3
Hence sufficient!
IMO B












