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

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If x and y are integers greater than 1, is x a multiple of [#permalink]

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16 Nov 2005, 02:56
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If x and y are integers greater than 1, is x a multiple of y?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-integers-greater-than-1-is-x-a-multiple-of-y-126736.html
[Reveal] Spoiler: OA
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26 Jul 2011, 21:28
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i think A.

Yalephd wrote:
If X and Y are integers greater than 1, is X a multiple of Y?

$$(1) 3Y^2 + 7Y = X$$

$$(2) X^2 - X$$ is a multiple of $$Y.$$

1st statement - Y(3Y+1) = X implies that... X is a multiple of Y... - sufficient

2nd Statement - i am not sure the best way... but i tried to use some arbitrary smart #s which will give me X>1.
X(X-1) is multiple of Y... so ex- if Y=6, then X =3.. so X is not a multiple of Y.
or Y = 12, then X = 4 again X is not a multiple of Y.

so not sufficient.
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26 Jul 2011, 22:13
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The answer should be (A) - statement (1) alone is sufficient to answer the question.

Consider statement (1) first:
X = 3Y^2 + 7Y
=> X = Y (3Y + 7)
=> X = Y * some number
Therefore X is a multiple of Y. So statement (1) alone is sufficient.

Next consider statement 2:
X^2 - X is a multiple of Y
=> kY = X(X-1) where k is any integer >= 1
=> X = kY/(X-1) = [k/(X-1)]Y
Therefore X is a multiple of Y depending on whether k/(X-1) is an integer or not.
For k = 4 and X = 5, this works. For k=5, X=5 it does not work.
Therefore statement (2) alone is insufficient to answer the question.

Hence (A) is the correct answer choice.
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16 Nov 2005, 03:55
themagiccarpet wrote:
If x and y are 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) x=3y^2+7y= y ( 3y+7) , since y is an integer---> 3y+7 is an integer ---> x is a multiple of y
---->suff

(2) x^2-x is a multiple of y
pick number x = 3 ,y =6 ---> x^2-x= 6 is a multiple of 6 but x=3 isn't
x=3, y=3, x^2-x= 6 is a multiple of y=3 , x=3 is a multiple of 3
--> insuff.
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16 Nov 2005, 14:55
D
If x and y are integers greater than 1, is x a multiple of y?
(1)3y^2 + 7y = x

x = y(3y+7) hence x is always a multiple as y is an integer so 3y+7 will also be an ingeter . Ans Yes .... Suff

(2)x^2 -x is a multiple of y
y=m*(x-1)x
as (x-1) > 1
and x is also an integer so Y is a multiple of x not the reverse.
No ...suff
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16 Nov 2005, 16:12
This was in the GMATPrep test I just took. It's A. Why? I have no clue.
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16 Nov 2005, 17:44
Ok lets see...

X and Y are positiv integers greater than 1

x>1, y>1

(1) can be written as y(3y+1)=x
y.m=x
as we know that y is an integer....3Y+1 is also an integer thus X is a multiple of Y...Sufficient

(2) x(x-1)=y
xm=y

hmm this lets see if x=4, y=2...then 4(3)=12...works fine and x is a multiple of Y...if x=5 y=4 then 5(4)=20=which is a multiple of y however x=5 is not a multiple of y=4

A it is
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DS - Prep software- divisibility [#permalink]

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30 Jan 2006, 14:03

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

1) 3y^2 + 7y = x

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

Last edited by allabout on 30 Jan 2006, 15:20, edited 1 time in total.
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30 Jan 2006, 15:11
A

x is a multiple of y means

x = ay where "a" is a +ve integer. (Because x and y are greater than 1)

St1: 3y^2 + 7y = x

i.e x = y(3y+7) ... so a = 3y+7 and will always be a +ve integer. So x is a multiple of y. SUFF

St2: x^2 - x... there is no information about y. INSUFF
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30 Jan 2006, 15:27
added another infromation, which doesn't affect the result.

Thank you dahiya

I did:

3y^2 + 7y = x

x/y = (3y+7)

which is essentally the same, but I wasn't sure if it's an proper way; of course it is but what made me doubt was

x^2 -x is a multiple of y

[x(x-1)]/y at first I wasn't sure if I can infer x/y and (x-1)/y, now I see clear.
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17 Feb 2006, 13:28
If x and y are 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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17 Feb 2006, 13:41
D.
from i, y (3y+7) = x, (3y+7) is an integer. so it is enough
from ii, x(x-1) = yk where k is an integer because x and (x-1) both are integer.
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18 Feb 2006, 00:55
Professor wrote:
D.
from i, y (3y+7) = x, (3y+7) is an integer. so it is enough
from ii, x(x-1) = yk where k is an integer because x and (x-1) both are integer.

briliant explanation))my approach was way too long
agree D it is
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18 Feb 2006, 01:06
Professor wrote:
D.
from i, y (3y+7) = x, (3y+7) is an integer. so it is enough
from ii, x(x-1) = yk where k is an integer because x and (x-1) both are integer.

A

professor will k/(x-1) always be an integer ...?
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18 Feb 2006, 02:08
k/(x-1) will not always be an integer, and that's the point. So X^2 - X is a multiple of y does not make necessarily x a multiple of y.

Try x=4 y=3
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18 Feb 2006, 02:15
agree with D)
to Professor, think that for B) it should be Y=K(X^2-X) or Y=KX(X-1) and YES X is a multiple of Y
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18 Feb 2006, 04:19
Agree with (B)

x(x-1)/y -> either x or (x-1) is multiple of y, or both
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18 Feb 2006, 05:27
k/(x-1) will not always be an integer, and that's the point. So X^2 - X is a multiple of y does not make necessarily x a multiple of y.

Try x=4 y=3

stmt1, holds good

but stmt 2 doesnt

tryx=3, then x(x-1) = 6 = y

x is not a multiple of y

try x=2, then y=2
x is a multiple of Y

I'm with A
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18 Feb 2006, 06:32
tryx=3, then x(x-1) = 6 = y

x is not a multiple of y

Do you really think that 3 is not a multiple of 6?
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18 Feb 2006, 07:04
gmacvik wrote:
If x and y are integers greater than 1, is x a multiple of y?

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

let me correct myself and change to A.

what i overlooked with (ii) is I assumed (X^2 - X) is a multiple of y, which is given. ststement (ii) doesnot tell that x is a multiple of y. if suppose x = 5 and y = 2.

(X^2 - X) = yk
25-5=2k
k=10.

but x=5 is not a multiple of y=2. so ii is not suff.
Re: DS: Multiple   [#permalink] 18 Feb 2006, 07:04

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