## If x and y are integers greater than 1

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### If x and y are integers greater than 1

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.

The answer is A. Not sure how.

RonPurewal
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Joined: Tue Aug 14, 2007 8:23 am

since this problem is about factors and multiples, you can gain considerable insight into it by factoring the polynomials.

(1)
y(3y + 7) = x
-- since y times some integer** is x, it follows that x is a multiple of y.
(** we know that 3y + 7 is an integer, because y is an integer)
sufficient

(2)
x(x - 1) is a multiple of y
-- could mean any of 3 things:
possibility a) x is a multiple of y
possibility b) x - 1 is a multiple of y
possibility c) neither x nor x - 1 is a multiple of y, but together they contain all the prime factors of y (e.g., x = 3, x - 1 = 2, y = 6)
insufficient, because we don't know whether it's possibility (a) or not

there you go (no need to test them together ... thankfully, as that'd be horrendous)
ajaykraju
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### Re: If x and y are integers greater than 1

Why is B a right choice?

x(x-1)=y X (some no.)
x= y X (some no.)/(x-1)
In this case also, x is a multiple of y. Isnt it?
RonPurewal
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### Re: If x and y are integers greater than 1

ajaykraju wrote:Why is B a right choice?

It's not. The answer is A (statement 1 is sufficient, but statement 2 is not).

x(x-1)=y X (some no.)
x= y X (some no.)/(x-1)
In this case also, x is a multiple of y. Isnt it?

Only if the purple thing is an integer. The purple thing isn't necessarily an integer.
RonPurewal
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### Re: If x and y are integers greater than 1

You can also quickly prove statement 2 insufficient with specific numbers. For instance, if x = 3, then x^2 - x = 6, which is a multiple of both 2 and 3. therefore, y could be either 2 or 3.
If x = 3 and y = 3, then y is a multiple of x.
If x = 3 and y = 2, then y is not a multiple of x.
Not sufficient.
anniee624
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### Re: If x and y are integers greater than 1

Ron, I am not quite understand what it means "X is multiply of y"

If X=2 Y=6,then X=1/2 * 6, then I think that X is a multiply of y. Then I think that all number can be multiply of ant other numbers.....

I think that I am definitely wrong in this thought..But I don't know whatÂ´s the right explain of it? If one interger A is multiply of another integer B, then A=n*B, n has to be an integer too?