hi earnest10
i guest you have problem with 1 st let`s look at it
x,y,z are +ve integers
x is a factor of y, it means that y=a*x where a is +ve integer
y is a factor of z , z=y*m where m is +ve integer
is z even
(1) z*x=even, 3 cases are possible
x,z both are even ( z even)
z even, x odd (again z even)
z odd x even- let`s prove that it is impossible
if x even, then y=ax, from here y even, and z=y*m , but as we proved that y is even z must be even too
so case: z odd x even here does not exist
z is always even

+1 for option D.

If x is a factor of y, which in turn is a factor of z, it follows that z is a multiple of y, which in turn is a multiple of x.
Note that if either x or y is even, z, which is a multiple of both, will also be even.

(1) If xz is even, either z or x is even, and if x is even, so is z ! SUFF
Alternatively, if z were odd, so would be x and thus xz would be odd as well, contradicting (1)

(2) If y is even, so is z, which is a multiple of y SUFF

Try to think about factors in the sense that they're DNA. So if x is a factor of y, then x is in the DNA of y. Same goes for z. If y is a factor of z, then y is part of z (and therefore so is x).

If you want, you can think about it in terms of real numbers. 2 is a factor of 4 because there are two 2s in 4. 4 is a factor of 12 because there is a 4 in 12. And so 2 is also a factor of 12, since 2 a factor of 4.

So now we're looking to see if z is even. You could rephrase that and ask, can we determine if x, y, or z are even? If one of them is, they all are.

Statement 1: xz = even. Well if the product of both is even, that means either x is, in which case so is z. OR, maybe x is 3 and z is 12... but z is still even. Therefore, sufficient.

Statement 2: y is even. Since we know y is in the DNA of z, we can clearly say that z is even, as well. Therefore, sufficient.

D

earnest10 wrote:The positive integers x, y and z are such that x is a factor of y and y is a factor of z. Is z even.

1) xz is even
2) y is even.

Let's reverse the language:

z is a multiple of y, and y is a multiple of x.
Thus, z is a multiple of x.

Statement 1: xz is even.
If x = even, then z must be even since it's a multiple of an even value.
If x = odd, then z must be even in order for xz to be even.
Since in either case z is even, sufficient.

Statement 2: y is even.
If z is a multiple of an even number, then z too must be even.
Sufficient.

The correct answer is D.