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

Question

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.

IanStewart · Answer

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.

BN1989 · Answer

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?

IanStewart · Answer

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.

siddharthmuzumdar · Answer

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.

shinbhu · Answer

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.

Bunuel · Answer

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  to get  wz=x+z ;

Re-arrange and factor out z to get  z(w-1)=x . Since  w>x>z  then  z=1  (if  z>1  then  x>w  which contradicts given condition);

Plug z = 1 into  z(w-1)=x  and re-arrange to get  w=x+1 .

w=x+1  means that  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 ;

w>x>w-2  (substituting  y  in the given inequality);

x=w-1 . The same as above. Sufficient.

Answer: D.

Hope it helps.

gmacforjyoab · Answer

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

-Jyothi

mau5 · Answer

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 eq{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

stne · Answer

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 
eq 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 
ot> 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.

PiyushK · Answer

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.

SVaidyaraman · Answer

What is tested is whether one is able to identify a specific set of consecutive integers.

Statement 1: z has to be 1 because w/x is greater than 1. z is the smallest integer and greater than 0. If z=1, we can see that w=x+1. A number cannot be a common divisor of two consecutive integers unless it is 1. y is not 1. Sufficient.

Statement 2: We can see W=y+2 and so w=x+1. Now the same reasoning as that for statement 1 applies. Sufficient.

s111 · Answer

GMATinsight  - please share your way of solving this question

Posted from my mobile device

GMATinsight · Answer

s111  Please find the solution as attached

sujoykrdatta · Answer

We know: w > x > y > z > 0, all are integers
We need to check if y is a common divisor of w and x

Statement 1: w/x = 1/z + 1/x 
Multiply x throughout:
=> w = x/z + 1
Multiply z throughout
=> wz = x + z
=> x = z(w - 1)

The value of w is at least 4. Thus, w-1 is at least 3
If z is 2 or more, x will exceed w
Thus, z = 1 
=> x = w-1
Thus, w and x are consecutive numbers => Their common factor is 1 
However, y is not 1 (since z = 1) => y is NOT a common divisor of w and x
Thus, statement 1 is sufficient

Statement 2: w^2 - wy - 2w = 0
=> w(w - y - 2) = 0
Since w is not 0, we have: w - y - 2 = 0
=> w - y = 2
=> w, x, y are consecutive numbers => Their common factor is 1 
However, y cannot be 1 (since z cannot be an integer) => y is NOT a common divisor of w and x
Thus, statement 2 is sufficient

Answer D

Ashu9876 · Answer

Sol.
Statement (1): w=x/z+1 . 
With respect to the above conditions we can try few numbers for w, x, y and z. 
W = 4, x = 3, y = 2, z = 1 
W = 6, x = 5, y = 3, z = 1
Hence with above of conditions there will be no set of positive integers in which y is common divisor of w and x. Hence statement 1 is sufficient. 
Statement (2): w^2-wy-2y=0,w-y=2 
With respect to the above conditions we can try few numbers for w, x, y and z. 
W = 4, x = 3, y = 2, z = 1 
W = 6, x = 5, y = 4, z = 3 
W, x, and y are consecutive. 
Hence with above of conditions there will be no set of positive integers in which y is common divisor of w and x. Hence statement 2 is sufficient. 
Both statements alone are sufficient to answer the questions.

JYnaey · Answer

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

---> is y a common divisor of w and x ? 
 if w and x are consecutive integers, the question is sufficient. We don't even have to know what y is.
 so, in a way (90%) it seems this question is asking whether w and x are consecutive or not.

* When a question mentions "common divisors", first thing that is springing in my mind is "Primes(co-prime)", next "Consecutive numbers" *

(1)  \frac{w}{x}= z^{-1}+x^{-1}

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

(1) w/x = 1/z + 1/x
 w = x/z +1 
 w-1 (integer) = x/z (must be integer)

if we plug-in values,

1,2,3 (z,x,w) ---> 3-1 = 2/1 (sufficient) 
 2,3,4 (z,x,w) ---> 4-1 = 3/2 ( we can see that from the moment z is more than 1, equation NOT valid.) 
 
 therefore Z must be 1. 
 w = x +1 (consecutive) =====> (1) is SUFFICIENT.

(2) same solution as above.

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

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