w, x, y, and z are integers

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guerrero: Master | Next Rank: 500 Posts; Posts: 227; Joined: Sun Apr 08, 2012 4:53 amFollowed by: 3 members; Thanked: 12 times

by guerrero » Fri Jul 26, 2013 12:37 pm

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

OA D[spoiler][/spoiler]

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Source: Beat The GMAT — Data Sufficiency | Fri Jul 26, 2013 12:37 pm

vishugogo: Master | Next Rank: 500 Posts; Posts: 131; Joined: Tue Aug 30, 2011 4:50 amFollowed by: 6 members; Location: India; Thanked: 28 times

by vishugogo » Fri Jul 26, 2013 2:17 pm

Statement 1:

w/x = 1/z + 1/x

because w > x

w/x is >1

1/x < 1

so z=1

so equations becomes

w/x = 1 + 1/x

w=x+1

w and x are consecutive integers so their common divisor can only be 1

If y=1 then z becomes zero which could not be the case

so y is not a common divisor

Statement 2:

w-y-2=0 (factor out a w)

so w=y+2

hence w=x+1

w and x are consecutive integers so their common divisor can only be 1

If y=1 then z becomes zero which could not be the case

so y is not a common divisor

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vishugogo: Master | Next Rank: 500 Posts; Posts: 131; Joined: Tue Aug 30, 2011 4:50 amFollowed by: 6 members; Location: India; Thanked: 28 times

by vishugogo » Fri Jul 26, 2013 2:18 pm

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Fri Jul 26, 2013 4:00 pm

Statement 1:

Let's simplify first.
w/x = 1/x + 1/z
w/x = (x+z)/xz
w = (x+z)/z
wz = (x+z)
wz - z = x
z(w-1) = x
(w-1) = x/z

Since w > x and w and x are integers, (w - 1) cannot be less than x. If z is an integer, x/z will be less than x UNLESS z = 1. So (w - 1) = x, and z = 1.

If x = (w - 1), then x and w are consecutive integers, and hence only have ONE common divisor: the number 1 itself. Since z = 1, and y > z, y CANNOT be a common divisor of x and w. SUFFICIENT!

Statement 2 is much easier to work with:

w^2 - wy - 2w = 0

Since w is greater than 0, divide both sides by w.

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

Since w and y are integers, and w > x > y, x must therefore be (w - 1). At this point we reach the same conclusion that we did in Statement 1, and this is statement is also SUFFICIENT.

This is a pretty sophisticated question, and it strikes me as something you'd only see if you're scoring Q49+ on the actual exam.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Jul 27, 2013 6:05 am

guerrero 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

A helpful rule to know:
If x and z are positive integers, and x/z is an integer, then x/z is a FACTOR OF X.
Proof:
Let x/z = k, where k is an integer.
Thus:
x = k * z
The resulting equation indicates that k -- and thus x/z -- is a factor of x.

Since w>x>y>z>0, the least possible values are w=4, x=3, y=2, and z=1.

Statement 1: w/x = 1/z + 1/x
Multiplying each side by x, we get:
w = x/z + 1.

The resulting equation implies that w = (factor of x) + 1.
If x=3, its factors are 1 and 3.
Since w must be 1 more than either 1 or 3, and w>x, the only option is w = 3+1 = 4.
If x=6, its factors are 1, 2, 3 and 6.
Since w must be 1 more than 1, 2, 3, or 6, and w>x, the only option is w = 6+1 = 7.

The cases above illustrate that w and x are CONSECUTIVE INTEGERS.
Consecutive integers are COPRIMES: they share no factors other than 1.
Since y>1, and w and x share no factors other than 1, it is not possible that y is a factor of both w and x.
SUFFICIENT.

Statement 2: wÂ²- wy- 2w = 0
w(w-y-2) = 0.
Since wâ‰ 0, it must be true that w-y-2 = 0.
Thus:
w = y+2.

If y=2, then w=4.
Since x must be an integer BETWEEN y and w, x=3.
The result:
w=4, x=3, y=2.

The case above indicates that w and x are CONSECUTIVE INTEGERS.
Since w and x are consecutive integers -- and thus share no factors other than 1 -- it is not possible that y is a factor of both w and x.
SUFFICIENT.

The correct answer is D.

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Anaira Mitch: Master | Next Rank: 500 Posts; Posts: 235; Joined: Wed Oct 26, 2016 9:21 pmFollowed by: 5 members; Thanked: 3 times

by Anaira Mitch » Tue Dec 27, 2016 1:57 am

GMATGuruNY wrote:
guerrero 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
A helpful rule to know:
If x and z are positive integers, and x/z is an integer, then x/z is a FACTOR OF X.
Proof:
Let x/z = k, where k is an integer.
Thus:
x = k * z
The resulting equation indicates that k -- and thus x/z -- is a factor of x.

Since w>x>y>z>0, the least possible values are w=4, x=3, y=2, and z=1.

Statement 1: w/x = 1/z + 1/x
Multiplying each side by x, we get:
w = x/z + 1.

The resulting equation implies that w = (factor of x) + 1.
If x=3, its factors are 1 and 3.
Since w must be 1 more than either 1 or 3, and w>x, the only option is w = 3+1 = 4.
If x=6, its factors are 1, 2, 3 and 6.
Since w must be 1 more than 1, 2, 3, or 6, and w>x, the only option is w = 6+1 = 7.

The cases above illustrate that w and x are CONSECUTIVE INTEGERS.
Consecutive integers are COPRIMES: they share no factors other than 1.
Since y>1, and w and x share no factors other than 1, it is not possible that y is a factor of both w and x.
SUFFICIENT.

Statement 2: wÂ²- wy- 2w = 0
w(w-y-2) = 0.
Since wâ‰ 0, it must be true that w-y-2 = 0.
Thus:
w = y+2.

If y=2, then w=4.
Since x must be an integer BETWEEN y and w, x=3.
The result:
w=4, x=3, y=2.

The case above indicates that w and x are CONSECUTIVE INTEGERS.
Since w and x are consecutive integers -- and thus share no factors other than 1 -- it is not possible that y is a factor of both w and x.
SUFFICIENT.

The correct answer is D.

Great Solution Mitch. Easily Understandable.

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hazelnut01: Master | Next Rank: 500 Posts; Posts: 157; Joined: Sat Nov 19, 2016 5:34 amFollowed by: 4 members; Thanked: 2 times

by hazelnut01 » Tue Mar 28, 2017 12:43 am

Matt@VeritasPrep wrote:Statement 1:

Let's simplify first.
w/x = 1/x + 1/z
w/x = (x+z)/xz
w = (x+z)/z
wz = (x+z)
wz - z = x
z(w-1) = x
(w-1) = x/z

Since w > x and w and x are integers, (w - 1) cannot be less than x. If z is an integer, x/z will be less than x UNLESS z = 1. So (w - 1) = x, and z = 1.

If x = (w - 1), then x and w are consecutive integers, and hence only have ONE common divisor: the number 1 itself. Since z = 1, and y > z, y CANNOT be a common divisor of x and w. SUFFICIENT!

Statement 2 is much easier to work with:

w^2 - wy - 2w = 0

Since w is greater than 0, divide both sides by w.

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

Since w and y are integers, and w > x > y, x must therefore be (w - 1). At this point we reach the same conclusion that we did in Statement 1, and this is statement is also SUFFICIENT.

This is a pretty sophisticated question, and it strikes me as something you'd only see if you're scoring Q49+ on the actual exam.

Dear @Matt@VeritasPrep, Could you help to elaborate the statement highlighted in red? I could not image the number.

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Tue Mar 28, 2017 3:42 am

ziyuenlau wrote:
Matt@VeritasPrep wrote:Statement 1:

Let's simplify first.
w/x = 1/x + 1/z
w/x = (x+z)/xz
w = (x+z)/z
wz = (x+z)
wz - z = x
z(w-1) = x
(w-1) = x/z

Since w > x and w and x are integers, (w - 1) cannot be less than x. If z is an integer, x/z will be less than x UNLESS z = 1. So (w - 1) = x, and z = 1.

If x = (w - 1), then x and w are consecutive integers, and hence only have ONE common divisor: the number 1 itself. Since z = 1, and y > z, y CANNOT be a common divisor of x and w. SUFFICIENT!

Statement 2 is much easier to work with:

w^2 - wy - 2w = 0

Since w is greater than 0, divide both sides by w.

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

Since w and y are integers, and w > x > y, x must therefore be (w - 1). At this point we reach the same conclusion that we did in Statement 1, and this is statement is also SUFFICIENT.

This is a pretty sophisticated question, and it strikes me as something you'd only see if you're scoring Q49+ on the actual exam.
Dear @Matt@VeritasPrep, Could you help to elaborate the statement highlighted in red? I could not image the number.

Matt was invoking the following axiom: two consecutive integers cannot have any divisors in common aside from 1.

To illustrate, take two random consecutive numbers, 15 and 16

Factors of 15: 1, 3, 5, 15
Factors of 16: 1, 2, 4, 8, 16

Notice, they have no factors in common, aside from 1. This makes sense. If 15 has 3 as a factor, the next number that will have 3 as a factor is 18. If 15 has 5 as a factor, the next number that will have 5 as a factor is 20. So we're continually leaping over that 16.

As a consequence, if I asked you if variable 'y' was a common divisor of 15 and 16, you'd know that the only way this could be true would be if y = 1. right? Similarly, if y is not equal to 1, then y can't be a common divisor of 15 and 16.

Same idea here. If x = w - 1, then x and w are consecutive integers. (In my scenario, w =16 and x = 15.) If z = 1 and y > z, then we know that y is NOT equal to 1. If y is NOT equal to 1, and x and w are consecutive integers, then y cannot be a divisor of those numbers, as two consecutive numbers have no factors in common, aside from 1.

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