-
sandranjeim
- Newbie | Next Rank: 10 Posts
- Posts: 8
- Joined: Sun Apr 18, 2010 3:05 am
- Thanked: 2 times
From 1:
X=2n+1 and X=3m, where m and n are integers. So basically X takes values which are odd multiples of 3 e.g. 3,9,15 etc
when divided by 6 remainder is always 3
SUFFICIENT
FROM 2:
X=12n+3 So basically X take values like 15, 27 etc (12n is always divisible by 6 leaving 3 as remainder)
when divided by 6 remainder is always 3
SUFFICIENT
IMO answer D












