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Multiples A

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Multiples A

by tanyajoseph » Mon Sep 10, 2007 12:56 pm
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

Elaborate explanation would be great :)
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TT

by TT » Mon Sep 10, 2007 3:29 pm
I think the answer is A.
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by ri2007 » Mon Sep 10, 2007 7:15 pm
Agree ans should be A

Stat 1) 3y^2+7y = x

3y^ is a multiple of Y, also 7 Y is a multiple of Y so the sum of the two will be a multiple of Y. Since Multiple of Y is = X ans should be yes.

You can plug in diff values and check.

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

Since x^2 - x can be greater than greater than X you cannot determine if X is also a multiple

I have a feeling I have not been able to explain this one properly. Hope it helps ????
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by samirpandeyit62 » Mon Sep 10, 2007 8:52 pm
Stat 1: 3y^2+7y = x

i.e y(3y+7) =x

so x is a multiple of y

stmt 2: x^2 - x is a multiple of y

so x(x-1) is a multiple of y

so y is a factor of x(x-1)

now we cannot say that x is a multiple of y i.e here x may be a multiple of y or it may be x-1

So ans should be A
Regards
Samir
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by girish3131 » Mon Mar 15, 2010 12:13 am
I perfectly agree with th eexplanation that B is not sufficient from yr perspective...

now jus take this in this way ->

x(x-1)/y = k where k is an integer

now x/y = k / (x-1)

now x > 1(given) so take any value for x let say 5

x/y = 4k ie an integer...

so in this way we can say that y is an factor of x

hence B too sufficient....



wat u say..
Last edited by girish3131 on Mon Mar 15, 2010 12:36 am, edited 1 time in total.
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by gmatmachoman » Mon Mar 15, 2010 12:15 am
samirpandeyit62 wrote:Stat 1: 3y^2+7y = x

i.e y(3y+7) =x

so x is a multiple of y

stmt 2: x^2 - x is a multiple of y

so x(x-1) is a multiple of y

so y is a factor of x(x-1)

now we cannot say that x is a multiple of y i.e here x may be a multiple of y or it may be x-1

So ans should be A
good explanation.
St 1: we can clearly say X is a mulitple of Y.(i.e y(3y+7) =x )

St 2: It can be that X or (X-1) can be the multiple of Y.So Insufficient

IMO A
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by girish3131 » Mon Mar 15, 2010 12:18 am
no no...

my point is that B too suffiecient irrespective any value of X
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by gmatmachoman » Mon Mar 15, 2010 12:49 am
girish3131 wrote:no no...

my point is that B too suffiecient irrespective any value of X
How come?
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by kstv » Mon Mar 15, 2010 12:55 am
x and y are integers greater than 1, is x a multiple of y ?
can be express x = ky where k can be any constant
1) 3y²+7y = x y(3y+7) = x
means y multiplied by (3y+7) = x let (3y+7) = k as k can be any constant
yk = x Suff.
2) x²-x is a multiple of y
so x(x-1) = ky where k can be any constant
x = k/(x-1) *y as x > 1 ( all the while u wonder why this fact was mentioned.)
k / (x-1) will not always be an integer.
if it was given x < 1 eg. 1/2 then k/(x-1) = 2k then x= 2k.y then Suff
but now it's Insufficient.

IMO A
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by girish3131 » Mon Mar 15, 2010 1:11 am
kstv wrote:x and y are integers greater than 1, is x a multiple of y ?
can be express x = ky where k can be any constant
1) 3y²+7y = x y(3y+7) = x
means y multiplied by (3y+7) = x let (3y+7) = k as k can be any constant
yk = x Suff.
2) x²-x is a multiple of y
so x(x-1) = ky where k can be any constant
x = k/(x-1) *y as x > 1 ( all the while u wonder why this fact was mentioned.)
k / (x-1) will not always be an integer.
if it was given x < 1 eg. 1/2 then k/(x-1) = 2k then x= 2k.y then Suff
but now it's Insufficient.

IMO A
KSTV

acc to u , for which value of x , k / (x-1) is not an integer ?
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by kstv » Tue Mar 16, 2010 10:40 pm
girish3131 wrote:
kstv wrote:x and y are integers greater than 1, is x a multiple of y ?
Is the assumption I am making too sweeping
please point out the flaw .
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