Multiples Problem

[vongdn]

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

[ramannjit]

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

[Rahul@gurome]

(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.

[blaster]

(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?

[Rahul@gurome]

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.