Is a three-digit number xyz divisible by 7?

Good Question!
Given 3 digit number is xyz and can be written as 100*x + 10*y + z
{ Ex: 123 can be written as 100*1 + 10*20 + 1*3 = 123 }

STAT1
|2*z -10x -y| is divisible by 7
=> |2*z -10x - y| = 7k (where k is an integer)
=> 2z - 10x -y = 7r (where r can be positive or negative depending on the sign of 2z - 10x - y (doesn't really matter in this question till the time we know that its a multiple of 7))

multiply both sides by -10 we get

-20z + 100x + 10y = -70r
-21z + z + 100x + 10y = -70r (breaking -20z into -21z + z)
or, 100x + 10y + z = 21z - 70r
21z - 70r is a multiple of 7
=> 100x + 10y + z is a multiple of 7, hence divisible by 7
So, xyz is divisible by 7
So, SUFFICIENT

STAT2
z + 3y + 2x = 7k (where k is an integer)

multiply both sides by 50 we get
50z + 150y + 100x = 350k

49z + z + 140y+ 10y+ 100x = 350k
or, 100x + 10y + z = 350k - 140y - 49z

350k - 140y - 49z is divisible by 7
So, 100x + 10y + z is also divisible by 7
So, xyz is divisible by 7
So, SUFFICIENT

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Absolute Values