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Difficulty: 505-555 Level,   Multiples and Factors,                                       
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Re: If x, y, and z are positive integers such that x is a factor of y, and [#permalink]
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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 of y, and [#permalink]
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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 of y, and [#permalink]
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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 of y, and [#permalink]
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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


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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 of y, and [#permalink]
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 of y, and [#permalink]
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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 of y, and [#permalink]
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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.

GMAT assassins aren't born, they're made,
Rich
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Re: If x, y, and z are positive integers such that x is a factor of y, and [#permalink]
EMPOWERgmatRichC wrote:
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.

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


EMPOWERgmatRichC
Thank you for this explanation. To clarify, isn't it always best practice to check all of the answer choices when testing values? Or, are you able to get away with not checking all the choices when testing values for this specific type of question (no actual values are given)?

Thanks again!
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Re: If x, y, and z are positive integers such that x is a factor of y, and [#permalink]
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woohoo921 wrote:
EMPOWERgmatRichC wrote:
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.

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


EMPOWERgmatRichC
Thank you for this explanation. To clarify, isn't it always best practice to check all of the answer choices when testing values? Or, are you able to get away with not checking all the choices when testing values for this specific type of question (no actual values are given)?

Thanks again!


Hi woohoo921,

Yes - when TESTing VALUES, you would normally check all 5 answer choices (for whatever result you were looking for). Here though, with the way that the prompt is written, we'll know immediately what the correct answer is when we see it (it's the one that won't always be an integer). Once we find the one answer that fits this description, then we can stop working.

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

Contact Rich at: Rich.C@empowergmat.com
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Re: If x, y, and z are positive integers such that x is a factor of y, and [#permalink]
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.


I approached it differently and found the answer, but i am wondering if my methods was lucky this timeor if it is a good one.
I basically though that : there i a relation between x and y; there is a relation between x and z.
But i can't find relation between y and z so (y + z) / x can't be determined as an interger or not.
was i lucky or did i use a trcik that work 100% of the time?
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Re: If x, y, and z are positive integers such that x is a factor of y, and [#permalink]
I used x as 4, y as 8 and z as 2, and all the choices were integers. Can someone help with this? Bunuel KarishmaB
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Re: If x, y, and z are positive integers such that x is a factor of y, and [#permalink]
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sanyashah wrote:
I used x as 4, y as 8 and z as 2, and all the choices were integers. Can someone help with this? Bunuel KarishmaB

­
The question asks: "which of the following is NOT necessarily an integer?" All options except B are always integers. However, B might be an integer, for specific ser of numbers, and might not be for another. Hence, option B is  NOT necessarily an integer.
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Re: If x, y, and z are positive integers such that x is a factor of y, and [#permalink]
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sanyashah wrote:
I used x as 4, y as 8 and z as 2, and all the choices were integers. Can someone help with this? Bunuel KarishmaB

Further,the post here may help.
It explains the difference between Must be true and Could be true.
­
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