[email protected] 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) w² - wy - 2w = 0

TRICKY question! I couldn't think of a super nice approach, so please forgive the length of this solution.


Important Concept: If integer k is greater than 1, and if k is a factor (divisor) of N, then k is not a divisor of N+1
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1)
Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313

We can extend this concept to say: If integer k is greater than 1, then k cannot be a divisor of both N and N+1
In other words, if integer k is greater than 1, then k cannot be a common divisor of two consecutive integers.

Okay, now onto the question . . .

Target question: Is y a common divisor of w and x?

Given: w > x > y > z > 0

IMPORTANT: From this, we can conclude that y > 1, which means we may be able to apply the green rule above.

Statement 1: w/x = z^-1 + x^-1
So, w/x = 1/z + 1/x
Rearrange terms to get: w/x - 1/x = 1/z
Rewrite to get: (w - 1)/x = 1/z
Cross multiply to get: z(w - 1) = x
IMPORTANT: This equation tells us that z = 1. Here's why.
Since it's given that x < w, we can replace x with z(w - 1) to get: z(w - 1) < w
Now divide both sides by (w - 1) to get: z < w/(w-1)
w/(w-1) is just slightly bigger than 1, and since we're told that z is positive, it MUST be the case that z = 1.
If z = 1, and z(w - 1) = x, then we can see that w - 1 = x
Or we can write, w = x + 1
This tells us that w is 1 greater than x, which means x and w are consecutive integers.
If y > 1, and x and w are consecutive integers, then (by the green rule above), we can be certain that y is definitely not a common divisor of w and x
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: w² - wy - 2w = 0
Factor to get: w(w - y - 2) = 0
This means that EITHER w = 0 OR w - y - 2 = 0
w cannot equal 0 since we're told that w > x > y > z > 0
So, it MUST be the case that w - y - 2 = 0
Rearrange to get w - y = 2
In other words, w is 2 greater than y
Since we're told that w > x > y > z > 0, we know that x must lie between the values of w and y, which means x and w are consecutive integers.
If y > 1, and x and w are consecutive integers, then (by the green rule above), we can be certain that y is definitely not a common divisor of w and x
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

Hi shibsriz,

I agree with Brent that this IS a bit of a tricky question (in that the patterns behind it are not readily apparent). BUT there are variables, so you can TEST Values and likely discover what the patterns are...

We're given a TON of information to start off with: W,X,Y and Z are integers and W > X > Y > Z > 0
We have to follow ALL of these rules when we TEST Values...

We're asked IF W/Y and X/Y are BOTH integers? This is a YES/NO question.

Fact 1: W/X = 1/Z + 1/X

At first glance, it might be tempting to call this insufficient, but we have so much information that we should "play around" with this a bit and find out what it really tells us. Since we're adding fractions, I'm going to focus on the common denominator:

WZ/XZ = X/XZ + Z/XZ

Since the variables are positive integers, we can get rid of the denominators:

WZ = X + Z

Now, I'm going to TEST Values (keep in mind that W > X > Y > Z > 0)

W=4
X=3
Z=1
4(1) = 3 + 1
With these values, the only possible value for Y is...
Y=2
4/2 AND 3/2 are not BOTH integers, so the answer is NO

Trying to change the Z to a 2 causes an interesting problem...

W(2) = X + 2
The X would have to be BIGGER than the W, which is NOT allowed. So Z CANNOT = 2; thus Z MUST = 1 and Y must be AT LEAST 2.

With THAT restriction, we have:
W(1) = X + 1
This tells us that W will ALWAYS be 1 greater than X.
There are NO possible values of Y that will divide evenly into 2 consecutive numbers.
Thus, the answer to the question will ALWAYS be NO.
Fact 1 is SUFFICIENT

Fact 2: W^2 - WY - 2W = 0

TEST Values:
W = 4
Y = 2
X must = 3
4/2 and 3/2 are NOT both integers, so the answer to the question is NO.

W = 5
Y = 3
X must = 4
5/3 and 4/3 are NOT both integers, so the answer to the question is NO.

W = 6
Y = 4
X must = 5
6/4 and 5/4 are NOT both integers, so the answer to the question is NO.

It appears that the answer is consistently NO.
Fact 2 is SUFFICIENT.

Final Answer:D

GMAT assassins aren't born, they're made,
Rich