w, x, y, and z are integers. If w > x > y > z > 0, is y a co : GMAT Data Sufficiency (DS)
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# w, x, y, and z are integers. If w > x > y > z > 0, is y a co

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w, x, y, and z are integers. If w > x > y > z > 0, is y a co [#permalink]

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10 Dec 2011, 16:51
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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 = z^-1 + x^-1
(2) w^2 - wy - 2w = 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Jul 2013, 12:44, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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10 Dec 2011, 17:48
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ashiima wrote:
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 = z^-1 + x^-1
(2)w^2-wy-2w=0

This is a pretty awkwardly designed question. From Statement 1 we know:

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

Now we know from the stem that w > x, or, since w and x are integers, that w > x + 1. Substituting in the expression we just found for w, we have that

(x/z) + 1 > x + 1
x/z > x

If z is an integer, this could only be true if z = 1 (if z is bigger than 1, clearly the left side of the inequality above will be smaller than the right side). So we know that z=1, and plugging that into our expression for w, we know that

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

So w is 1 greater than x, and w and x are therefore consecutive integers. The Greatest Common Divisor of any two consecutive positive integers is *always* equal to 1. Since y cannot be equal to 1 (since y > x > 0, and x and y are integers, the smallest possible value of y is 2), y cannot be a common divisor of x and w. So Statement 1 is sufficient.

From Statement 2 we can factor out a w:

w^2-wy-2w=0
w(w - y - 2) = 0

For this product to be 0, one of the factors on the left side must be 0. We know that w is not zero, so w - y - 2 must be zero:

w - y - 2 = 0
w = y + 2

But if w > x > y, and each of these quantities are integers, then if w is exactly 2 greater than y, it must be that w, x and y are three consecutive integers. So again we know that w and x are consecutive integers, and just as we saw in Statement 1, it is impossible for y to be a common divisor of both.

So the answer is D, since each statement is sufficient to give a 'no' answer to the question.

I find it an awkward question because it tries to confuse the fundamental concept behind the question (GCD of consecutive integers is 1) by introducing distractions like negative exponents. That's not something you see in real GMAT questions. It also is a question where the statements are sufficient to give a 'no' answer, and such questions are very rare on the real GMAT - in most yes/no DS questions, if statements are sufficient, the answer is 'yes'. The basic concept in this question is difficult enough without it needing to be complicated further.
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11 Dec 2011, 03:27
IanStewart wrote:
ashiima wrote:
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 = z^-1 + x^-1
(2)w^2-wy-2w=0

This is a pretty awkwardly designed question. From Statement 1 we know:

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

Now we know from the stem that w > x, or, since w and x are integers, that w > x + 1. Substituting in the expression we just found for w, we have that

(x/z) + 1 > x + 1
x/z > x

If z is an integer, this could only be true if z = 1 (if z is bigger than 1, clearly the left side of the inequality above will be smaller than the right side). So we know that z=1, and plugging that into our expression for w, we know that

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

So w is 1 greater than x, and w and x are therefore consecutive integers. The Greatest Common Divisor of any two consecutive positive integers is *always* equal to 1. Since y cannot be equal to 1 (since y > x > 0, and x and y are integers, the smallest possible value of y is 2), y cannot be a common divisor of x and w. So Statement 1 is sufficient.

From Statement 2 we can factor out a w:

w^2-wy-2w=0
w(w - y - 2) = 0

For this product to be 0, one of the factors on the left side must be 0. We know that w is not zero, so w - y - 2 must be zero:

w - y - 2 = 0
w = y + 2

But if w > x > y, and each of these quantities are integers, then if w is exactly 2 greater than y, it must be that w, x and y are three consecutive integers. So again we know that w and x are consecutive integers, and just as we saw in Statement 1, it is impossible for y to be a common divisor of both.

So the answer is D, since each statement is sufficient to give a 'no' answer to the question.

I find it an awkward question because it tries to confuse the fundamental concept behind the question (GCD of consecutive integers is 1) by introducing distractions like negative exponents. That's not something you see in real GMAT questions. It also is a question where the statements are sufficient to give a 'no' answer, and such questions are very rare on the real GMAT - in most yes/no DS questions, if statements are sufficient, the answer is 'yes'. The basic concept in this question is difficult enough without it needing to be complicated further.

Why is it impossible for y to be a common divisor of both in statement 2, there's no restriction to the value of y given in statement 2, so y could be 1, which would make it a common divisor of w and x, wouldnt it?
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11 Dec 2011, 13:36
BN1989 wrote:

Why is it impossible for y to be a common divisor of both in statement 2, there's no restriction to the value of y given in statement 2, so y could be 1, which would make it a common divisor of w and x, wouldnt it?

There is a restriction on the value of y, in the question. We know that z and y are integers, and that y > z > 0. So the smallest possible value of z is 1, and the smallest possible value of y is 2. It is not possible, just from the stem alone, that y is 1 here.
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04 Jan 2012, 12:24
First of all, Ian, wonderful approach. I learned a lot from your approach. Thanks.

I followed the following crude approach to solve the problem:

We know that w > x > y > z > 0. Thus the least value of z is 1. Least value of y is 2.

I considered the values of z, y, x and w as
Case 1. z = 1, y = 2, x = 3, w = 4. In this case, y is NOT a factor of both x and w
Case 2. z = 1, y = 2, x = 4, w = 8. In this case, y is a factor of both x and w

Statement 1: Substitute the values for each case in the equation given (w/x) = (1/z) + (1/x)
Case 1: The values satisfy the equation.
Case 2: The values do not satisfy the equation.

Statement 2: w^2 - wy - 2w = 0 => w(w - y - 2) = 0 => either w = 0 or (w - y - 2) = 0.
w cannot be 0 because w > 0. Thus, w - y = 2.
Substitute the values for each case in the equation above.
Case 1: The values satisfy the equation.
Case 2: The values do not satisfy the equation.

As seen from both statements, only 1 case satisfies the given statements. Thus, the given numbers are consecutive integers. We can see that y is, thus, not a factor of both x and w.
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04 Jan 2012, 12:37
Tough one and calculation intensive - obviously being from MGMAT.

w > x > y > z > 0
are w/y or x/y integers

I am going to use AD/BCE strategy here. S2 appears simple..

2. Rephrase gives us w = 0, w = y+2. But w>0, so w = y+2.
That means that w,x,y are a sequence. There is no way y becomes factor of w and x. Sufficient
B removed, now D.

1. Rephrase gives wz = x+z. Lets plug numbers here
4,3,2,1 satisfies the above, y is not a factor of w and x
5,4,2,1 or 5,4,3,1 satisfies, but y is not a factor of w and x
Sufficient.

D it is.
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Re: MGMAT   [#permalink] 04 Jan 2012, 12:37
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