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w, x, y, and z are integers. If w > x > y > z > 0, is y a common

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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

Originally posted by ashiima on 10 Dec 2011, 17:51.
Last edited by Bunuel on 10 Jul 2013, 13:44, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 26 Jan 2012, 23:33
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enigma123 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


w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) \(\frac{w}{x}= z^{−1} + x^{−1}\) -> \(\frac{w}{x}=\frac{1}{z}+\frac{1}{x}\) --> multiply both sides by \(xz\) --> \(wz=x+z\) --> \(z(w-1)=x\) --> since \(w>x>z\) then \(z=1\) (if \(z>1\) then \(x>w\) which contradicts given condition) --> \(w=x+1\) --> \(w\) and \(x\) are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus \(w\) and \(x\) don not share any common factor more than 1 and as \(y>1\) (from \(y>z>0\)) then \(y\) is not a common factor of \(w\) and \(x\). Sufficient.

(2) \(w^2 – wy – 2w = 0\) --> \(w(w-y-2)=0\), since \(w>0\) then: \(w-y-2=0\) --> \(w-2=y\) -->s \(w>x>w-2\) (substituting \(y\) in the given inequality) --> \(x=w-1\). The same as above. Sufficient.

Answer: D.

Hope it helps.
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w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post Updated on: 23 Nov 2017, 08:06
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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) \(\frac{w}{x}= z^{-1}+x^{-1}\)

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

Folks - started to solving it like this, but got stuck after a while. Can someone please help?

Basically this question is asking whether y is factor of both w & x?

Considering Statement 1

w/x=1/z+1/x. Since w>x , the quotient w/x must be greater than 1. For example 17/6 = 2 5/6

Since w/x is greater than 1 then 1/z + 1/x must be greater than 1 as well. For 1/z+1/x to be greater than 1 both z and x has to be 1. x cannot be equal to 1 because of the condition w > x > y > z > 0, therefore, z must be equal to 1. We can now re write the equation as w/x = 1+1/x which will give us w = x+1. If w= x+1 then mean x and w are consecutive integers and therefore no number other than 1 can be a factor of both.

I got stuck after this. But from my guess work I think this statement looks sufficient. I know not the right way of doing MATHS, therefore can someone please help?

Considering Statement 2

w^2-wy-2w = 0

If we factor above into, that will give w(w-y-2) = 0. This means that either w = 0 or w-y-2 = 0. w cannot be zero because w > x > y > z > 0, therefore w-y-2 = 0 or w = y+2. Since x falls between y and w we can tell that x, y and w are consecutive integers and therefore no number other than 1 can be a factor of both. Since y >1 it cannot be a factor of both w & x. Therefore this statement is sufficient to answer the question.

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Originally posted by enigma123 on 26 Jan 2012, 22:19.
Last edited by Bunuel on 23 Nov 2017, 08:06, edited 5 times in total.
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New post 10 Dec 2011, 18:48
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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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Re: MGMAT  [#permalink]

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New post 11 Dec 2011, 04:27
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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New post 11 Dec 2011, 14:36
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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Re: MGMAT  [#permalink]

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New post 04 Jan 2012, 13:24
1
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.
Thus, D is the answer.
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Re: MGMAT  [#permalink]

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New post 04 Jan 2012, 13:37
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1
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: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 26 May 2013, 21:45
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enigma123 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



1 ) w/x = 1/z + 1/x
since W is greater than x , w/x should be greater than 1
hence 1/z +1/x should also be greater than 1. since x and z are integers grater than 0 , and z<x , the only way , 1/z +1/x can be >1 is if z=1
hence equation becomes - w/x= 1/1 + 1/x ======> w-1=x ..
hence x and w are consecutive integers and since y is not 1 and is >1 ( since z is 1) , the answer is NO
Sufficient

2) w^2-wy-2w=0
w(w-y-2) =0
since w != 0 w=y+2 .. w, x and y are consecutive integers . Same as statement 1 , the answer is NO.
Suff

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

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New post 27 May 2013, 00:21
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enigma123 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


From F.S 1, we know that \(\frac{w}{x}= z^{-1}+x^{-1}\) --> \(w = \frac{x}{z}+1\)

or \(w-1 = \frac{x}{z}\).

On the number line,we can have this arrangement ---x ----- x/z ---- w ; where \(\frac{x}{z}\) and w are consecutive integers.

Note that we can not have ---x/z---x---w as because there cannot be an integer(x) between 2 consecutive integers.

Thus, we have \(\frac{x}{z}>x\) --> z<1. However, this is not possible as z is at-least 1. Thus, the only solution is when \(\frac{x}{z}\) and x coincide i.e. \(\frac{x}{z}\) = x --> z=1.

As w and x are consecutive integers, y(which is not equal to 1) can never be a divisor for both x and w. Sufficient.

From F. S 2, we know that as\(w\neq{0}\) w = y+2. Thus, on the number line,

--y ---(y+1)----w. Thus, there is only one integer between y and w and as y<x<w, we have x = y+1. Just as above, y can't be a divisor for both w and x. Sufficient.


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

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New post Updated on: 08 Dec 2016, 07:21
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enigma123 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

Folks - started to solving it like this, but got stuck after a while. Can someone please help?

Basically this question is asking whether y is factor of both w & x?

Considering Statement 1

w/x=1/z+1/x. Since w>x , the quotient w/x must be greater than 1. For example 17/6 = 2 5/6

Since w/x is greater than 1 then 1/z + 1/x must be greater than 1 as well. For 1/z+1/x to be greater than 1 both z and x has to be 1. x cannot be equal to 1 because of the condition w > x > y > z > 0, therefore, z must be equal to 1. We can now re write the equation as w/x = 1+1/x which will give us w = x+1. If w= x+1 then mean x and w are consecutive integers and therefore no number other than 1 can be a factor of both.

I got stuck after this. But from my guess work I think this statement looks sufficient. I know not the right way of doing MATHS, therefore can someone please help?

Considering Statement 2

w^2-wy-2w = 0

If we factor above into, that will give w(w-y-2) = 0. This means that either w = 0 or w-y-2 = 0. w cannot be zero because w > x > y > z > 0, therefore w-y-2 = 0 or w = y+2. Since x falls between y and w we can tell that x, y and w are consecutive integers and therefore no number other than 1 can be a factor of both. Since y >1 it cannot be a factor of both w & x. Therefore this statement is sufficient to answer the question.



Awesome sum, here's my small bit


\(\frac{w}{x} = \frac{1}{z} +\frac{1}{x}\) , Given

\(\frac{w}{x} - \frac{1}{x}= \frac{1}{z}\), bringing terms with same denominator to LHS

\(\frac{w-1}{x}=\frac{1}{z}\), simplification

now if z>1 then RHS is a fraction so LHS is a fraction too.

Now if LHS is a fraction then x>w-1

lets look at this statement x>w-1, since w and x are positive and \(x \neq w\) hence x>w-1 is never possible.
Lets take a few positive integers such that w>x and test this. e.g. x = 6 and w =7 , 6 > 6 not possible, x= 6 w=12 , 6>11 not possible , hence x> w-1 not possible if w>x and positive.So LHS cannot be a fraction and hence RHS also cannot be a fraction so Z cannot be > 1


if Z>1 not possible then Z=1 and as \(0 < z \not> 1\) ( z is greater than 0 but not greater than 1 and z is an integer so z can be 1 only)

so if z = 1 then from \(\frac{w-1}{x}=\frac{1}{z}\) we have w-1 = x , which tells us that x is one less than w or x and w are consecutive integers. Hence they are co prime . Therefore they cannot have a common factor other than 1, but we know z is 1 so y cannot be 1, because y>z , Hence y is not a common factor for w and x as y is not 1 but >1.
Sufficient.
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 16 May 2014, 14:51
Another easy approach is to introduce y in the equation by inserting...w=py and x=qy
The equation formed after putting in the values in I is not valid & hence w & x do not have a common factor x ...as the equation cannot be untrue
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 17 May 2014, 00:49
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1
Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers.
And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2.
W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 01 Aug 2014, 16:10
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enigma123 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



(1) w = x/z + 1 -> x = kz and w = k + 1 ( with k is an positive integer)
since w > x -> k+1> kz -> k(z-1) < 1
because of k and z are positive integers, z must equal 1 -> x = k , and w = k + 1 -> y is not a common divisor of w and x.

(2) w = y + 2 -> x = y + 1 ..... -> y is not a common divisor of w and x.

answer D
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 03 Sep 2014, 14:54
PiyushK wrote:
Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers.
And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2.
W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.

@PiyushK - I think you meant "W and X must be consecutive numbers"... not "W and Z" :)
@Bunuel - In your solution, for st. 1, you wrote
Bunuel wrote:
since w>x>z then z=1 (if z>1 then x>w which contradicts given condition)
. How did you reach this conclusion?
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 03 Sep 2014, 18:35
ronr34 wrote:
PiyushK wrote:
Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers.
And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2.
W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.

@PiyushK - I think you meant "W and X must be consecutive numbers"... not "W and Z" :)
@Bunuel - In your solution, for st. 1, you wrote
Bunuel wrote:
since w>x>z then z=1 (if z>1 then x>w which contradicts given condition)
. How did you reach this conclusion?


w > x > y > z > 0, is y a common divisor of w and x?

We have that z(w-1)=x and w>x>z>0. If z>1, say if z=2, then 2(w-1)=x. Now tell me can in this case x be less than w?
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 28 Jul 2015, 11:46
1
Assuming w=4, x=3, y=2, z=1

St1:

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

Using the above values, Y is not a divisor of W & X. Sufficient.

St2 :

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

Hence, w= y + 2

using the above values, Y is not divisible of W & X. Sufficient.

Final ans D)
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 27 Nov 2015, 01:11
Given to us : w > x > y > z > 0

In such type of questions, we have to start by reducing the given equations to a simple form and then test the simplified equations with a random number, so to quickly come to a required answer.

Using Statement 1

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

Simplifying the equation

\(w/x=(x+z)/xz\)
\(wz=x+z\)
\((w-1)z=x\) (Simplified Equation)

Now from the given equation \(w>x>y>z>0\), we definitely know that z minimum will be equal to 1. Substituting z=1 in the simplified equation, we get
\(w=x+1\).
From this equation, we can thus deduce that w and x are consecutive integers and it also satisfies w>x criteria. So if two numbers are consecutive to each other there can only be one number common to them (because these numbers become co-prime to each other, example 4 & 5) i.e. 1. but we definitely know y cannot be 1. It will definitely be greater than 1 because z=1 and y>z. So, y is not a common divisor of x and w.
Now, in case if z>1, then from the simplified equation, we will have x>w, which will contradict the given condition of (w>x>y>z>0).
Hence, statement 1 is sufficient is sufficient to answer the question that y is not a common divisor of x and w.

Using Statement 2

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

Simplifying the equation

\(w(w-y-2)=0\)
So, either \(w=0\) or \(w-y-2=0\)

w=0 is not possible because \(w>x>y>z>0\) is given to us.

Considering, \(w-y-2=0\), we get
\(w=y+2\).

Using this equation, we can conclude that there will be an integer between w and y, and that is definitely going to be x, because of the equation w>x>y>z>0.
For example, if w=6, then y becomes 4, then x has to be 5 to satisfy the equation.

So, from this statement we definitely know, that w,x and y are three consecutive integers.
Now three consecutive integers can only be co-prime to each other and can never have any common divisor in between them. There can only be one number common in between these numbers i.e. 1

So from this statement we can conclude that y is not the common divisor in between x and w.
Hence, Statment 2 is also sufficient to answer the Question.

So the Answer is D.
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common  [#permalink]

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New post 13 Mar 2016, 23:05
Bunuel wrote:
ronr34 wrote:
PiyushK wrote:
Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers.
And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2.
W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.

PiyushK - I think you meant "W and X must be consecutive numbers"... not "W and Z" :)
Bunuel - In your solution, for st. 1, you wrote
Bunuel wrote:
since w>x>z then z=1 (if z>1 then x>w which contradicts given condition)
. How did you reach this conclusion?


w > x > y > z > 0, is y a common divisor of w and x?

We have that z(w-1)=x and w>x>z>0. If z>1, say if z=2, then 2(w-1)=x. Now tell me can in this case x be less than w?



I did not understand one concept here
What if W=-10
w-1=-11
z=100
=> z*(w-1) = -200
so x=-200
also x<w is satisfied..


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

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New post 13 Mar 2016, 23:28
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Re: w, x, y, and z are integers. If w > x > y > z > 0, is y a common &nbs [#permalink] 13 Mar 2016, 23:28

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