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Multiples Problem

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Source: — Data Sufficiency |
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by ramannjit » Wed Oct 06, 2010 9:05 pm
vongdn 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



ANSWER --> (1) alone is sufficient
You can find explaination to the above in this post :)

https://www.beatthegmat.com/is-x-a-multi ... 16147.html
Ramannjit
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by Rahul@gurome » Wed Oct 06, 2010 9:09 pm
(1) implies y(3y+7) = x.
Since y is an integer, 3y+7 is an integer.
Or x is a multiple of y.
So (1) alone is sufficient.

Next consider (2) alone.
Let x = 3 and y = 2.
So x^2 - x = 6 is a multiple of y = 2, but x = 3 is not a multiple of y =2.
Next let x = 4 and y = 2.
So x^2 - x = 12 is a multiple of y = 2 and x = 4 is a multiple of y = 2
We do not get a definite answer.
So (2) alone is not sufficient.
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by blaster » Thu Oct 07, 2010 2:56 am
Rahul@gurome wrote:(1) implies y(3y+7) = x.
Since y is an integer, 3y+7 is an integer.
Or x is a multiple of y.
So (1) alone is sufficient.

Next consider (2) alone.
Let x = 3 and y = 2.
So x^2 - x = 6 is a multiple of y = 2, but x = 3 is not a multiple of y =2.
Next let x = 4 and y = 2.
So x^2 - x = 12 is a multiple of y = 2 and x = 4 is a multiple of y = 2
We do not get a definite answer.
So (2) alone is not sufficient.
Dear Rahul
is there any algebraic method for statement (2) other than the number pick?
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by Rahul@gurome » Thu Oct 07, 2010 11:17 pm
Dear Rahul
is there any algebraic method for statement (2) other than the number pick?
If x^2 - x is a multiple of y, we can write that y divides x(x-1).
This means that there is a possibility that y divides (x-1) and y does not divide x.
In such a case x will not be a multiple of y.
However if y divides x(x-1), there is another possibility that y is dividing x.
In such a case x is a multiple of y.
Since nothing definite can be said (2) alone is not sufficient.
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