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Senior Manager  Joined: 24 Feb 2008
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Schools: UCSD ($) , UCLA, USC ($), Stanford
If x and y are both integers greater than 1, is x a multiple  [#permalink]

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1
8 00:00

Difficulty:   45% (medium)

Question Stats: 64% (01:32) correct 36% (01:53) wrong based on 278 sessions

### HideShow timer Statistics If x and y are both integers greater than 1, is x a multiple of y?

(1) 3y^2 + 7y = x
(2) x^2 -x is a multiple of y

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SVP  Joined: 04 May 2006
Posts: 1560
Schools: CBS, Kellogg
Re: DS: x multiple of y?  [#permalink]

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chineseburned wrote:
If x and y are both integers greater than 1, is x a multiple of y?

(1) 3y^2 + 7y = x
(2) x^2 -x is a multiple of y

Hi, it is long time to see you!

1 suff. no discustion
2. x(x-1) is multiple of y. So x may or may not is multiple of y.

a. x=5, y = 2
5*6 is multiple of 2, but x=5 is not multiple of 2

b. x=6
6*7 is multiple of 2, and x=6 is multiple of 2
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Posts: 1293
Re: DS: x multiple of y?  [#permalink]

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1
chineseburned wrote:
If x and y are both integers greater than 1, is x a multiple of y?

(1) 3y^2 + 7y = x
(2) x^2 -x is a multiple of y

A.

Given:
x > 1
y > 1
n, x, y = integer

(1) x = y*(3y + 7)
Because y is integer, (3y + 7) must be integer; therefore, x must equal integer * y
SUFFICIENT

(2) x^2 - x = ny
Plug in numbers to satisfy above condition...
Say x=3, n=1, then y=6. In this case, x is not a multiple of y.
Say x=6, n=15, then y=2. In this case, x is a multiple of y.
The solution actually depends on what n is, and the only condition we have is n is integer. Therefore, it is INSUFFICIENT
SVP  Joined: 29 Aug 2007
Posts: 2242
Re: DS: x multiple of y?  [#permalink]

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chineseburned wrote:
If x and y are both integers greater than 1, is x a multiple of y?

(1) 3y^2 + 7y = x
(2) x^2 -x is a multiple of y

(1) 3y^2 + 7y = x
y (3y + 7) = x
so x must be a multiple, (3y+7) times, of y. suff...

(2) x^2 -x is a multiple of y
x^2 - x = yk where k is an integer.
x (x-1) = yk
from this we do not know whether x - 1 or x is equal to k. so nsf....

A.
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Intern  Joined: 08 Apr 2008
Posts: 40
Re: DS: x multiple of y?  [#permalink]

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GMAT TIGER wrote:

(2) x^2 -x is a multiple of y
x^2 - x = yk where k is an integer.
x (x-1) = yk
from this we do not know whether x - 1 or x is equal to k. so nsf....

What is the reasoning behind multiplying y by another variable in Statement 2 (in this case k)?
Director  Joined: 14 Aug 2007
Posts: 635
Re: DS: x multiple of y?  [#permalink]

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AlinderPatel wrote:
GMAT TIGER wrote:

(2) x^2 -x is a multiple of y
x^2 - x = yk where k is an integer.
x (x-1) = yk
from this we do not know whether x - 1 or x is equal to k. so nsf....

What is the reasoning behind multiplying y by another variable in Statement 2 (in this case k)?

2)x^2 -x is a multiple of y

let K be the multiple.

then x^2-x = K . y

Hope this helps
Intern  Joined: 10 Jun 2010
Posts: 2
Re: DS: x multiple of y?  [#permalink]

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so if we have

y^2-y = X

Then we could say with certitude that X is a multiple of Y?
Intern  Joined: 02 May 2010
Posts: 45
Schools: IU, UT Dallas, Univ of Georgia, Univ of Arkansas, Miami University
WE 1: 5.5 Yrs IT
Re: DS: x multiple of y?  [#permalink]

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Yes. If it was given that y^2-y = x, then x is certainly a multiple of y.

y^2-y = x
y(y-1) = x
y*k = x
Intern  Joined: 16 Jul 2010
Posts: 17
Re: DS: x multiple of y?  [#permalink]

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So from x(x-1)=yk, can we not derive:

x=y*k/(x-1)=yk?

In which case the answer is C, not A.
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Math Expert V
Joined: 02 Sep 2009
Posts: 56251
Re: DS: x multiple of y?  [#permalink]

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1
dauntingmcgee wrote:
So from x(x-1)=yk, can we not derive:

x=y*k/(x-1)=yk?

In which case the answer is C, not A.

We don't know whether $$\frac{k}{x-1}$$ is an integer, hence we can not write $$x=yn$$ (where n is an integer) from $$x(x-1)=yk$$.

If x and y are integers great than 1, is x a multiple of y?

Is $$x=ny$$, where $$n=integer\geq{1}$$?

(1) $$3y^2+7y=x$$ --> $$y(3y+7)=x$$ --> as $$3y+7=integer$$, then $$y*integer=x$$ --> $$x$$ is a multiple of $$y$$. Sufficient.

(2) $$x^2-x$$ is a multiple of $$y$$ --> $$x^2-x=my$$ --> $$x(x-1)=my$$ --> $$x$$ can be multiple of $$y$$ ($$x=2$$ and $$y=2$$) OR $$x-1$$ can be multiple of $$y$$ ($$x=3$$ and $$y=2$$) or their product can be multiple of $$y$$ ($$x=3$$ and $$y=6$$). Not sufficient.

Hope it helps.
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Re: DS: x multiple of y?  [#permalink]

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Yes, that is great. Thank you. Guess I'm a little rustier than I thought.
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GMAT 1: 460 Q42 V14 Re: If x and y are both integers greater than 1, is x a multiple  [#permalink]

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could someone please explain me the option 2 by plugging numbers. I have not understood the above solutions.
Thank you.
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Re: If x and y are both integers greater than 1, is x a multiple  [#permalink]

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SonGoku wrote:
could someone please explain me the option 2 by plugging numbers. I have not understood the above solutions.
Thank you.

(2) $$x^2−x$$ is a multiple of y

Number plugging approach:
Let $$y=2$$
If $$x=3$$ --> $$x^2−x$$ = $$3^2-3 = 6$$ --> $$x^2−x$$ is a multiple of y BUT x is not a multiple of y --> NO
If $$x=4$$ --> $$x^2−x$$ = $$4^2-4 = 12$$ --> $$x^2−x$$ is a multiple of y and x is a multiple of y --> YES
--> Insufficient.

Hope it's clear.
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