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# w, x, y, and z are integers. If w > x > y > z >

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Director
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w, x, y, and z are integers. If w > x > y > z > [#permalink]  20 May 2007, 05:35
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w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1)w/x = (1/z) + (1/x)
(2) w^2 - wy - 2w = 0

OA later
Manager
Joined: 11 Mar 2007
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Is it C? If so, let me know and I'll give my reasoning. If not, I don't want to look like an idiot...lol
Director
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I don't want to tell OA now. Let others apply their minds.
Senior Manager
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For me it is B.
w-y-2=0
w=y+2
hence
x=y+1
so: y, x and w are consecutives.
And they are not 1,2 and 3 as there is still z which is greater than 0.
y is not a common divisor of w and x
Manager
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divisibility [#permalink]  20 May 2007, 10:05
i found the same answer with the same thinking as Caas. I think B.
Manager
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I'm gonna go with D on this one.

For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.

If you look at statement 1:

w/x = (1/z) + (1/x)

Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.

So now, we have:

w/x = 1 + (1/x)

Multiply both sides by x and we get:

w = x + 1. Therefore, w and x are consecutive integers.

Statement 2:

w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.

Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.

Therefore, D?
CEO
Joined: 29 Mar 2007
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Im lost
Manager
Joined: 12 Apr 2007
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just curious -- how realistic is it to get one of these types of questions on the actual GMAT?

I have no idea how I would be able to figure out a problem like this in under 2 minutes and I'd probably end up guessing.

This problem took me a good 10 minutes to get a good grasp of, so I'm pretty scared now haha
Manager
Joined: 23 Mar 2007
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plaguerabbit wrote:
I'm gonna go with D on this one.

For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.

If you look at statement 1:

w/x = (1/z) + (1/x)

Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.

So now, we have:

w/x = 1 + (1/x)

Multiply both sides by x and we get:

w = x + 1. Therefore, w and x are consecutive integers.

Statement 2:

w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.

Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.

Therefore, D?

I too thought on the same lines. It has to be D . Can we have the OA please ?
Director
Joined: 14 Jan 2007
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plaguerabbit wrote:
I'm gonna go with D on this one.

For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.

If you look at statement 1:

w/x = (1/z) + (1/x)

Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.

So now, we have:

w/x = 1 + (1/x)

Multiply both sides by x and we get:

w = x + 1. Therefore, w and x are consecutive integers.

Statement 2:

w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.

Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.

Therefore, D?

PERFECT explanation. OA is 'D'
Director
Joined: 14 Jan 2007
Posts: 779
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Kudos [?]: 71 [0], given: 0

plaguerabbit wrote:
I'm gonna go with D on this one.

For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.

If you look at statement 1:

w/x = (1/z) + (1/x)

Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.

So now, we have:

w/x = 1 + (1/x)

Multiply both sides by x and we get:

w = x + 1. Therefore, w and x are consecutive integers.

Statement 2:

w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.

Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.

Therefore, D?

PERFECT explanation. OA is 'D'
Manager
Joined: 11 Mar 2007
Posts: 69
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Kudos [?]: 2 [0], given: 0

Makes sense now. My error was that in statement 1, i did not catch the fact they were consecutive int's. Glad to have you guys to fall back on.
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