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If x, y, and z are positive integers such that x is a factor

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If x, y, and z are positive integers such that x is a factor  [#permalink]

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x

Problem Solving
Question: 172
Category: Arithmetic Properties of numbers
Page: 85
Difficulty: 600


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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 18 Mar 2014, 01:36
4
4
SOLUTION

If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x

Given: \(z\) goes into \(x\) and \(x\) goes into \(y\). Note that it's not necessarily means that \(z<x<y\), it means that \(z\leq{x}\leq{y}\) (for example all three can be equal x=y=z=1);

Now, in all options but B we can factor out the denominator from the nominator and reduce it. For example in A: \(\frac{x+z}{z}\) as \(z\) goes into \(x\) we can factor out it and reduce to get an integer result (or algebraically as \(x=zk\) for some positive integer \(k\) then \(\frac{x+z}{z}=\frac{zk+z}{z}=\frac{z(k+1)}{z}=k+1=integer\)).

But in B. \(\frac{y+z}{x}\) we can not be sure that we'll be able factor out \(x\) from \(z\) thus this option might not be an integer (for example x=y=4 and z=2).

Answer: B.

Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice.
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 18 Mar 2014, 03:54
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Answer = (B) \(\frac{(y+z)}{x}\)

Took x = 6; y = 12; z = 2

6 is a factor of 12, and 6 is a multiple of 2

Only option B contradicts the condition
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 19 Mar 2014, 22:20
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IMO B.
let y = 10
then x = 2 or 5.
z can be 1 or 2

B satisfies.
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 20 Mar 2014, 00:14
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1
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x

Problem Solving
Question: 172
Category: Arithmetic Properties of numbers
Page: 85
Difficulty: 600


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

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We have integers Y,X and Z multiples of Z, all not equal to 0 and Y>=X>=Z

(A) (x+z)/z: we add multiples of Z and divide it by Z, it is always integer
(B) (y+z)/x: it is integer when Y=X=Z and Y>X=Z and not integer when Y>X>Z and Y=X>Z
(C) (x+y)/z: the same as in A
(D) (xy)/z: if one multiple of other, multiplying with any other guarantees multiple
(E) (yz)/x: the same as in D

For test strategy we could use picking but it has danger for testing B. We can choose B by exluding ACDE
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 23 Mar 2014, 08:11
1
1
SOLUTION

If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x

Given: \(z\) goes into \(x\) and \(x\) goes into \(y\). Note that it's not necessarily means that \(z<x<y\), it means that \(z\leq{x}\leq{y}\) (for example all three can be equal x=y=z=1);

Now, in all options but B we can factor out the denominator from the nominator and reduce it. For example in A: \(\frac{x+z}{z}\) as \(z\) goes into \(x\) we can factor out it and reduce to get an integer result (or algebraically as \(x=zk\) for some positive integer \(k\) then \(\frac{x+z}{z}=\frac{zk+z}{z}=\frac{z(k+1)}{z}=k+1=integer\)).

But in B. \(\frac{y+z}{x}\) we can not be sure that we'll be able factor out \(x\) from \(z\) thus this option might not be an integer (for example x=y=4 and z=2).

Answer: B.

Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice.
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Collection of Questions:
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 06 Apr 2014, 02:19
A: z is obviously divisible by z and x is div by z, since x is a multiple of z (x = zn). So, x+z is div by z
B: y is div by x, since x is a factor of y. z is not div by x if x>z. for example: z=6 x=12. So, y+z is not necessarily div by x.
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 06 Apr 2014, 02:42
[quote="Bunuel"]The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x

If we plug in: x = 6, y = 12 and z = 3.

A - (9)/3
B - (15)/12 - NOT an integer
C - (18)/3
D - Integer
E - Integer

Hence answer will be B
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 02 Jun 2015, 02:22
3
1
Given
We are given 3 positive integers \(x\), \(y\) and \(z\) such that \(x\) is a factor of \(y\) and \(x\) is a multiple of \(z\). We are asked to find that among the options given which of them is not necessarily an integer.

Approach
Since \(x\) is a factor of \(y\), we can say that \(\frac{y}{x}\) is an integer...........(1)

Also, as \(x\) is a multiple of \(z\) i.e. \(z\) is a factor of \(x\), we can say that \(\frac{x}{z}\) is an integer............(2)

From the above two deductions, we can say that \(\frac{y}{z}\) will also be an integer as \(x\) divides \(y\) completely and \(z\) divides \(x\) completely..............(3)

Our endeavor would be to reduce the expressions in the options to one or more of the above 3 forms.

Working Out
(A) \(\frac{x +z}{z} = \frac{x}{z} + 1\) . Since \(\frac{x}{z}\) is an integer, the expression will be an integer

(B) \(\frac{y+z}{x} = \frac{y}{x} + \frac{z}{x}\) . \(\frac{y}{x}\) is an integer but we can't say if \(\frac{z}{x}\) is also an integer. So the expression need not necessarily be an integer.

Although we have got our answer, I am reducing the other expressions for solution purpose.

(C) \(\frac{x+y}{z} = \frac{x}{z} + \frac{y}{z}\). Both \(\frac{x}{z}\) and \(\frac{y}{z}\) are integers, hence the expression will also be an integer

(D) \(\frac{xy}{z}\). Since \(\frac{x}{z}\) is an integer, the expression will also be an integer.

(E) \(\frac{yz}{x}\). Since \(\frac{y}{x}\) is an integer, the expression will also be an integer.

Hence, answer is Option B

Hope this helps :)

Regards
Harsh
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 03 Jun 2015, 16:16
1
Hi All,

This question is perfect for TESTing VALUES. Since the prompt asks 'which of the following is NOT necessarily an integer?", we can take advantage of a great 'shortcut' - we just need to find ONE instance for any answer to NOT be an integer and we'll have the solution...

We're given a few facts to work with:
1) X, Y and Z are POSITIVE INTEGERS
2) X is a FACTOR of Y
3) X is a MULTIPLE of Z

Let's TEST VALUES.....

IF....
X = 2
Y = 2
Z = 1

Answer A: (X+Z)/Z = (2+1)/1 = 3 This is an integer
Answer B: (Y+Z)/X = (2+1)/2 = 3/2 This is NOT an integer, so this MUST be the answer. We don't even have to check the others.

Final Answer:

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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 03 Jun 2015, 21:09
1
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x

Problem Solving
Question: 172
Category: Arithmetic Properties of numbers
Page: 85
Difficulty: 600


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!


An easy approach would be to bring everything in terms of z.

x = za (where a is a positive integer)
y = xb = zab (where b is a positive integer)

(A) (x+z)/z = (za + z)/z = a + 1 (positive integer)
(B) (y+z)/x = (zab + z)/za = (ab + 1)/a (Not necessarily an integer)
(C) (x+y)/z = (zab + za)/z = ab + a (positive integer)
(D) (xy)/z = zazab/z = zaab (positive integer)
(E) (yz)/x = zabz/za = zb (positive integer)

Answer (B)
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 04 Jun 2015, 00:33
B
As x is factor y and z is a factor of x that means z is a factor of y too
Only B divides z/x which might not be an integer
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 27 Aug 2015, 06:57
z x y
z 3z 9z --> (B) (y+z)/x = 9z+z/3z=10z/3z
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 22 Nov 2015, 21:05
I decided to pick numbers with exponents. Y=2^3 X=2^2 Z=2

All the ones with Z on the bottom are obviously divisible with my numbers and Y is stated to be divisible by x

Had to check B) 10/4 Not an integer.
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 24 Jan 2017, 08:02
Bunuel wrote:
SOLUTION

If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?

(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x

Given: \(z\) goes into \(x\) and \(x\) goes into \(y\). Note that it's not necessarily means that \(z<x<y\), it means that \(z\leq{x}\leq{y}\) (for example all three can be equal x=y=z=1);

Now, in all options but B we can factor out the denominator from the nominator and reduce it. For example in A: \(\frac{x+z}{z}\) as \(z\) goes into \(x\) we can factor out it and reduce to get an integer result (or algebraically as \(x=zk\) for some positive integer \(k\) then \(\frac{x+z}{z}=\frac{zk+z}{z}=\frac{z(k+1)}{z}=k+1=integer\)).

But in B. \(\frac{y+z}{x}\) we can not be sure that we'll be able factor out \(x\) from \(z\) thus this option might not be an integer (for example x=y=4 and z=2).

Answer: B.

Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice.


My opinion is that in GMAT terms the right solution is ONLY the:

"Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice
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If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 23 Jan 2018, 11:08
Hey everyone,

this question took me 7 minutes to solve correctly, is there any shortcut for dummies to solve this quckly ? :? :)

here is my solution :) its only 30sec read, please read it below you wont get bored :)

Let y be 6
Then x = 2 or x =3
Z =1 or z =2

x is a factor of y -----> 2 is factor of 6
x is a multiple of z -----> 2 is multiple of 1

so let`s test them; this is the most exciting part of problem solving process :)


(A) (x+z)/z ( 2+2)/2 or (2+1)1
(B) (y+z)/x (6+1)/2= non integer or (6+2)/2 integer
(C) (x+y)/z 2+6/1 or 2+6/2
(D) (xy)/z 2*6/1 or 3*6/2
(E) (yz)/x 6*1/2 , 6*1/2 6*2/3 etc 
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 23 Jan 2018, 11:58
1
dave13 wrote:
Hey everyone,

this question took me 7 minutes to solve correctly, is there any shortcut for dummies to solve this quckly ? :? :)

here is my solution :) its only 30sec read, please read it below you wont get bored :)

Let y be 6
Then x = 2 or x =3
Z =1 or z =2

x is a factor of y -----> 2 is factor of 6
x is a multiple of z -----> 2 is multiple of 1

so let`s test them; this is the most exciting part of problem solving process :)


(A) (x+z)/z ( 2+2)/2 or (2+1)1
(B) (y+z)/x (6+1)/2= non integer or (6+2)/2 integer
(C) (x+y)/z 2+6/1 or 2+6/2
(D) (xy)/z 2*6/1 or 3*6/2
(E) (yz)/x 6*1/2 , 6*1/2 6*2/3 etc 


Hi dave13

In fact what your method i.e. substitution works best for this question and is the fastest route. try \(z=1\), \(x=2\) & \(y=4\). remember from question stem you should have realized that \(y>x>z\), hence you should pick the smallest number for \(z\) for easy calculation and then work upwards.

Algebraic approach -

given \(y=kx\) (where \(k\) is a constant) and \(x=qz\) (where \(q\) is a constant). In the options I could see that only \(x\) & \(z\) are in the denominators so

\(\frac{y}{x}=k=integer\) and \(\frac{x}{z}=q=integer\). with this understanding scan the options

A) \(\frac{(x+z)}{z}=\frac{x}{z}+\frac{z}{z}=integer+1=integer\)

(B) \(\frac{(y+z)}{x}=\frac{y}{x}+\frac{z}{x}=integer+non-integer=non-integer\)

(C) \(\frac{(x+y)}{z}=\frac{x}{z}+\frac{y}{z}=integer+integer=integer\)

(D) \(\frac{(xy)}{z}=y*\frac{x}{z}=y*integer=integer\)

(E) \(\frac{(yz)}{x}=\frac{y}{x}*z=integer*z=integer\)

Hence our answer is B
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Re: If x, y, and z are positive integers such that x is a factor  [#permalink]

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New post 23 Jan 2018, 13:20
Hi niks18,

You have to be careful about the assumptions you make about the given information. At NO point does the prompt state that Y > X.

In my approach, I used X=2, Y=2, Z=1 and found the correct answer WITHOUT having to check 3 of the options.

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Re: If x, y, and z are positive integers such that x is a factor &nbs [#permalink] 23 Jan 2018, 13:20
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