If x, y and z are integers and xyz is divisible by 8, is x e

Just think in terms of prime factors;

xyz is divisible by 8.
means; x*y*z must have at least 3 2's, doesn't matter from where those 2 come; the 3 2's are there.

Then think what possible ways can I get those three 2's.
x=8, y=1,z=1
We got 3 2's; this time from x.

x=1,y=8,z=1
I got 3 2's; this time from y.

x=1,y=1,z=8
I got 3 2's; this time from z.

There are many such combinations.
x=2,y=2,z=2
x=4,y=2,z=1
x=1,y=2,z=4

Q: Is x=even?

1. yz is divisible by 4.
This means that
yz must have at least two 2's because 4 has two 2's.
Now,
if y=2, z=2; yz=4;
But from the stem we know xyz contains three 2's
Thus, the one additional 2 must come from x AND x becomes even. If "x" contains one 2, it becomes even.

But, we don't know whether y=2, z=2;
y may be 4 or 8 or 16
Then the odd/even for x or z doesn't make a difference. Because, y alone took care of both statements.

If y=8; y has 3 2's
Now even if x=1; z=1; xyz will be divisible by 8 AND yz will be divisible by 4.
Thus, we saw two different scenario where x may be an even or an odd.
Not sufficient.


2. x,y and z are all NOT divisible by 4.
This statement tells us that neither of x, y or z is divisible by 4. What does it mean?

It means;
x does not contain two 2's. It may contain 0 2's or 1 2's.
y does not contain two 2's. It may contain 0 2's or 1 2's.
z does not contain two 2's. It may contain 0 2's or 1 2's.

However, we already know that xyz contain 3 2's.
Combining both the stem condition and statement 2, we can conclude that each variable has one AND only one 2 in its factors.

x=2
y=2
z=2

OR

x=2*3*7*11*13*13*13
y=2*5*43
z=2

The point is x will contain exactly one 2 in its factor. And if x contains one 2 in its factors, it must be even.
Sufficient.

Ans: "B"