Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

Given: x is a factor of y --> \(y=mx\), for some non-zero integer \(m\); y is a factor of z --> \(z=ny\), for some non-zero integer \(n\); So, \(z=mnx\).

Question: is z even? Note that \(z\) will be even if either \(x\) or \(y\) is even

(1) \(xz\) even --> either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.

(2) \(y\) even --> as \(z=ny\) then as one of the multiples of z even --> z even. Sufficient.

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.

Please explain. thanks.

y/x = k where k is an integer. y = xk ....................i

z/y = m where m is an integer. z = ym = xkm .....................ii

If a factor is even, then the source of the factor must be even.

1) If xz is even, z must be even because x may or may not be an even because x is a factor of z but z must be even. SUFF. 2) If y is even, z must be even because y is a factor of z. SUFF..

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.

Please explain. thanks.

Though Bunuel has provided the solution, I would just like to bring to your notice a train of thought.

When we say, "The positive integers x, y, and z are such that x is a factor of y and y is a factor of z.", it implies that if x or y is even, z will be even.

e.g. x = 4. Since x is a factor of y, y will be a multiple of 4 and will be even. Since z is a multiple of y, it will also be even. So, in a way, the 2 in x will drive through the entire sequence and make everything even.

Once this makes sense to you, it will take 10 secs to arrive at the solution. Stmnt 1: Either x or z (or both which will happen if x is even) is even. In either case, z is even. Stmnt 2: y is even, so z must be even
_________________

Bunuel, you always try to solve the questions algebraically, don't you?

Not at all. There are certain GMAT questions which are pretty much only solvable with plug-in or trial and error methods (well at leas in 2-3 minutes). Also many questions can be solved with logic and common sense much quicker than with algebraic approach. So you shouldn't always rely on algebra. Having said that I must add that there are of course other types of questions which are perfect for algebraic approach, plus I often use algebra just to explain a solution.
_________________

I like to thing of the boxes method. If you draw them out, then x is inside y which is inside z. zx, a 2 will exist inside the box of either z or x (which is itself inside z) so YES y a 2 will exist inside the box of y which is itself z so YES

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.

Please explain. thanks.

Though Bunuel has provided the solution, I would just like to bring to your notice a train of thought.

When we say, "The positive integers x, y, and z are such that x is a factor of y and y is a factor of z.", it implies that if x or y is even, z will be even.

e.g. x = 4. Since x is a factor of y, y will be a multiple of 4 and will be even. Since z is a multiple of y, it will also be even. So, in a way, the 2 in x will drive through the entire sequence and make everything even.

Once this makes sense to you, it will take 10 secs to arrive at the solution. Stmnt 1: Either x or z (or both which will happen if x is even) is even. In either case, z is even. Stmnt 2: y is even, so z must be even

awesome. your explanations and bunuel as well, are amazing thanks
_________________

Re: The positive integers x, y, and z are such that x is a [#permalink]

Show Tags

24 Feb 2012, 10:46

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. Please explain. thanks.[/quote]

Ans. let us take any 3 numbers, say x=3,y=18,z=54, or x=2,y=4,z=20 1)if xz is even then it means that either x or z is even,say that x is even, now there is no even number which is a factor of odd number, so z is definitely even, now if x is odd as in the above case, still then we can point out that z is even. 2)if y is even then it is clear that z will be even. Thus this question could be answered by any of the two questions.

if we plug and play, why can't we test X = 1? then y/n..

Plug in method would be far more painful for this question (and most other questions in my opinion). Think how you would go about it: Checking whether stmnt 1 is enough: xz is even If x = 1, y = 1 and z = 2 (so that xz is even), then z is even. If x = 2, y = 2 and z = 4, z is again even. Then you start thinking if you can take some values such that xz is even but z is not... Now you start using logic... Wouldn't you say it is far better to use logic in the first place itself?
_________________

(1) \(xz\) even --> either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.

Can you please provide a numerical example for this part it isn't very clear. Thanks in advance!

I thought it was insufficient as there were multiple cases either X or Y even or both....
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

(1) \(xz\) even --> either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.

Can you please provide a numerical example for this part it isn't very clear. Thanks in advance!

I thought it was insufficient as there were multiple cases either X or Y even or both....

We know that x is a factor of \(y \to y = Ix\) Again, \(z = yI^'\). Now, given that xz - even.

Case I:Assume that x = even , z = odd. Now, as x is even, y = even(I can be odd/even,doesn't matter). Again, as y is even, z HAS to be even(\(I^'\) is odd/even, doesn't matter). Thus, if x is even, z IS even. Numerical Example :y = 2*I(x=2). \(z = 6*I^'\). z IS even.

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.

Given: x is a factor of y --> \(y=mx\), for some non-zero integer \(m\); y is a factor of z --> \(z=ny\), for some non-zero integer \(n\); So, \(z=mnx\).

Question: is z even? Note that \(z\) will be even if either \(x\) or \(y\) is even

(1) \(xz\) even --> either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.

(2) \(y\) even --> as \(z=ny\) then as one of the multiples of z even --> z even. Sufficient.

Answer: D.

Perfect explanation. Remember the factor foundation rule. Also, other properties of factors that might be helpful to have in mind.

Just to remind you, The factor foundation rule states that "if a is a factor of b, and b is a factor of c, then a is a factor of c" Also, if 'a' is a factor of 'b', and 'a' is a factor of 'c', then 'a' is a factor of (b+c). In fact, 'a' is a factor of (mb + nc) for all integers 'm' and 'n' If 'a' is a factor of 'b' and 'b' is a factor of 'a', then 'a=b' If 'a' is a factor of 'bc' and gcd (a,b) = 1, then 'a' is a factor of 'c' If 'p' is a prime number and 'p' is a factor of 'ab' then 'p' is a factor of 'a' or 'p' is a factor of 'b' In other words,, any integer is divisible by all of its factors- and it is also divisible by all of the factors of its factors

Re: The positive integers x, y, and z are such that x is a [#permalink]

Show Tags

30 Jan 2015, 14:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: The positive integers x, y, and z are such that x is a [#permalink]

Show Tags

05 Feb 2016, 17:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...