The positive integers x, y, and z are such that x is a : GMAT Data Sufficiency (DS)
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# The positive integers x, y, and z are such that x is a

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

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17 Oct 2009, 13:15
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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Feb 2012, 01:35, edited 1 time in total.
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amitgovin 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.

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
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 36568 Followers: 7082 Kudos [?]: 93232 [7] , given: 10553 Re: positive integers [#permalink] ### Show Tags 17 Oct 2009, 13:32 7 This post received KUDOS Expert's post 14 This post was BOOKMARKED amitgovin 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. 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. _________________ Math Expert Joined: 02 Sep 2009 Posts: 36568 Followers: 7082 Kudos [?]: 93232 [2] , given: 10553 Re: positive integers [#permalink] ### Show Tags 12 Dec 2010, 09:48 2 This post received KUDOS Expert's post metallicafan wrote: 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. _________________ SVP Joined: 29 Aug 2007 Posts: 2492 Followers: 67 Kudos [?]: 734 [1] , given: 19 Re: positive integers [#permalink] ### Show Tags 19 Oct 2009, 22:18 1 This post received KUDOS amitgovin 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. 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.. D.. _________________ Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT Moderator Joined: 01 Sep 2010 Posts: 3089 Followers: 785 Kudos [?]: 6525 [1] , given: 1012 Re: positive integers [#permalink] ### Show Tags 26 Dec 2010, 16:37 1 This post received KUDOS VeritasPrepKarishma wrote: amitgovin 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. 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 _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13660 [1] , given: 222 Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 27 Feb 2012, 04:20 1 This post received KUDOS Expert's post dchow23 wrote: 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? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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03 Aug 2013, 03:26
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fozzzy wrote:
Bunuel wrote:

(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.

Case II : z is even OR (x and z) both are even.

Hope this helps.
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12 Dec 2010, 09:22
Bunuel, you always try to solve the questions algebraically, don't you?
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12 Dec 2010, 18:36
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
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17 Mar 2011, 22:07
Easy if you realize the following:
When a is a factor of b
AND b is a factor of c
THEN a is a factor of c as well.

Hence when either one of these numbers is even, the other has to be even too..
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17 Mar 2011, 22:52
z = ky

y = mx

so z = (km)xy

(1) -> xz is eve means at least x or z is even, and if x = even, then z is also even as it has an even factor.

2 -> y is even so z having an even factor is even too.

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15 Apr 2011, 04:24
At first, mistook "factor" for "multiple" came with answer E.
Later, understood that the problem was so easy...just plug and play !!
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Re: The positive integers x, y, and z are such that x is a [#permalink]

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24 Feb 2012, 09:06
if we plug and play, why can't we test X = 1? then y/n..
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Re: The positive integers x, y, and z are such that x is a [#permalink]

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24 Feb 2012, 09:46
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.

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

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22 Jan 2013, 21:49
amitgovin 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.

What is given?

$$z = y*N$$
$$y = x*R$$

1.
$$xz = 2*I$$
If x is even, then z is even since $$z = x*R$$
If z is even, then z is even.
SUFFICIENT!

2. $$y = 2*I$$ ==> $$z = 2*I*N$$ Definitely EVEN
SUFFICIENT!

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03 Aug 2013, 01:04
Bunuel wrote:

(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....
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10 Oct 2013, 14:01
Bunuel wrote:
amitgovin 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.

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.

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

Hope it helps
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