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

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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

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