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

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chineseburned
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If x and y are both integers greater than 1, is x a multiple [#permalink] New post   04 May 2008, 19:26
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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
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A
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Re: DS: x multiple of y? [#permalink] New post   17 Jul 2010, 08:12
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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, 20:35
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, 20:55
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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


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, 21:29
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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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Re: DS: x multiple of y? [#permalink] New post   24 Jun 2008, 18:02
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, 23:26
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, 07:41
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, 07:54
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, 07:52
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, 10:46
Yes, that is great. Thank you. Guess I'm a little rustier than I thought.
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Re: If x and y are both integers greater than 1, is x a multiple [#permalink] New post   19 Aug 2018, 00:29
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] New post   19 Aug 2018, 09:12
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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Re: If x and y are both integers greater than 1, is x a multiple [#permalink] New post   23 Oct 2020, 18:20
For those confused with statement #2, follow this approach:

Statement #1) = 3y^2+7y=x -> y(3y+7)=x -> (y=2;x=26;Yes) (y=3;x=48;Yes) (y=4;x=76;Yes) -> Sufficient
Statement #2) = x^2-x is a multiple of y -> x(x-1) is a multiple of y
-> If x=3 then x(x-1)=6 so y is a factor of 6. (y=2;No) (y=3;Yes) -> Not Sufficient

Hope that helps.
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Re: If x and y are both integers greater than 1, is x a multiple [#permalink] New post   17 Jan 2024, 18:02
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