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If x and y are both integers greater than 1, is x a multiple

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If x and y are both integers greater than 1, is x a multiple [#permalink] New post 04 May 2008, 18:26
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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
[Reveal] Spoiler: OA

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Re: DS: x multiple of y? [#permalink] New post 17 Jul 2010, 07:12
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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.

Answer: A.

Hope it helps.
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Re: DS: x multiple of y? [#permalink] New post 04 May 2008, 19:35
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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Re: DS: x multiple of y? [#permalink] New post 04 May 2008, 19:55
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
Asking: Is x = ny?

(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
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Re: DS: x multiple of y? [#permalink] New post 04 May 2008, 20:29
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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Re: DS: x multiple of y? [#permalink] New post 24 Jun 2008, 17:02
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)?
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Re: DS: x multiple of y? [#permalink] New post 24 Jun 2008, 22:26
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
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Re: DS: x multiple of y? [#permalink] New post 10 Jun 2010, 06:41
so if we have

y^2-y = X

Then we could say with certitude that X is a multiple of Y?
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Re: DS: x multiple of y? [#permalink] New post 10 Jun 2010, 06:54
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
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Re: DS: x multiple of y? [#permalink] New post 17 Jul 2010, 06:52
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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Re: DS: x multiple of y? [#permalink] New post 17 Jul 2010, 09:46
Yes, that is great. Thank you. Guess I'm a little rustier than I thought.
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Re: DS: x multiple of y?   [#permalink] 17 Jul 2010, 09:46
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