The positive integers x, y and z are such that x

[nr]

From GMAT prep :

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.

Answer is D.

Can someone help me with this.


Thanks.

[pravsr]

Hi,

I think the best way to solve this problem is to plug in different numbers.

1) xz = even.

Take x=3 y=6 and z=12. xz=even.. z is even
Take x=2 y=4 and z=8. xz=even.. z is even

Say xz is not even. Take x=1, y=3 and z=9.. z is not even.

So, sufficient.

2) y is even

An even integer will have all its multiples as even. Since z is a multiple of y, z has to be even.

So, sufficient.

D is the right answer

[RonPurewal]

You can use the 'prime box' concept here.

Question 'Is z even?' --> 'Does the prime box for Z contain the number 2?'

(2) is the easy statement. If y is even, then y has a 2 in its prime factorization. But if y is a factor of z, then all the prime factors of y are also in z. Therefore, z contains a '2' in its prime box, so Z is even. Sufficient.

(1) means that either x is even, z is even, or both.
* If z is even, then the answer to the question is an immediate Yes.
* If x is even, then x contains a '2' in its prime factorization. Since x is a factor of y, so does y. Since y is a factor of z, so does z. So z is even. Sufficient.

Either is sufficient, so, D.

[sudaif]

Ron - if i were to solve this question using the basic even-odd divisibility properties, that should suffice right? do let me know if following thought process is ok. thanks.


Rephrased question
Per the factor foundation rule, x must be a factor of z b/c it is a factor of y and y is factor of z.
Question is, is z even?

statement 1:
X * Z = even
E * E = even  z is even and x is even. This works b/c z would then also be divisible by x.
E * O = even  does not work b/c z which is odd in this case, will not be divisible by any even number. So not possible
O * E = even  this will work. There are plenty of even numbers that are divisible by odd numbers.

Statement 2:
If Y = even, then z must be even b/c no odd number is divisible by an even number.

[RonPurewal]

Ron - if i were to solve this question using the basic even-odd divisibility properties, that should suffice right? do let me know if following thought process is ok. thanks.


Rephrased question
Per the factor foundation rule, x must be a factor of z b/c it is a factor of y and y is factor of z.
Question is, is z even?

statement 1:
X * Z = even
E * E = even  z is even and x is even. This works b/c z would then also be divisible by x.
E * O = even  does not work b/c z which is odd in this case, will not be divisible by any even number. So not possible
O * E = even  this will work. There are plenty of even numbers that are divisible by odd numbers.

Statement 2:
If Y = even, then z must be even b/c no odd number is divisible by an even number.

yep.
by definition, anything that works by considering "multiple of 2 vs. not multiple of 2" will also work by considering even vs. odd, since those are two ways of describing exactly the same properties.

the key aspect of this problem is realizing that x is a factor of y, which is also a factor of z -- and, therefore, anything that goes into either x or y must also go into z.