If x and y are integers, what is the value of xy? (1) x^2 - y^2

Question

If x and y are integers, what is the value of xy?

(1)  x^2 - y^2 < 0

(2)  x^y = \frac{1}{16}

Tulkin987 · Answer

Statement (1) clearly is not sufficient, because x and y can take multiple integer values.

If we solve statement (2) for   integer values   of x and y, we result in  (16; -1) ,  (4; -2)  and  (2; -4) . Still, not sufficient.

If we take both statements together, only  (2; -4)  satisfies given conditions. Hence  x*y=2*-4=-8 .

Answer: C

NiharikaR · Answer

St1:
(x-y)(x+y)<0
insufficient

St 2
(2)^-4, (-2)^-4, (4)^-2, (-4)^-2, 16^-1
multiple values of xy 
insufficient

Combining
x=+2,-2 correspondingly y=-4
xy=+8,-8

IMO E

aviejay · Answer

Bunuel

Hi Bunuel, can you please explain where am i going wrong?

From 1, we get -y<x<y
From 2, we get, 4^-2, 2^-4 and 16^-1

We can discard 4^-2 and 16^-1 as clearly they do not satisfy 1
Now, in 2^-4, x=2 and y=-4. So -y=4. So here too we see that 4!<2!<-4, that is -y!<x!<y. Hence 1 is not satisfied though I understand that 1 essentially means x lies between y and -y.
Though 2^-4 satisfies 1 logically, it does not do so mathematically. Hence, the answer should be E and not B

Please explain .

ColumbiaEMBA · Answer

1 is insufficient by itself, you're right there.
For 2, consider the following:

x and y can take on the following values to equal \frac{1}{16}: (2, -4), (-2, -4), (16, -1), (4, -2). This makes 2 by itself insufficient.

If you combine 1 and 2: 
 2^2 - (-4)^2  = 4 - 16 = -12
 (-2)^2 - (-4)^2  = 4 - 16 = -12

You have 2 possible answer which makes C insufficient. Correct answer therefore must be E.

ColumbiaEMBA · Answer

The correct answer i s  E. Also, don't forget  (-4)^{-2}

ruchik789 · Answer

You said there are two possible answers but the question asked was x*y which is -12 always. Hence why not C. Did i forget some other case

ColumbiaEMBA · Answer

X *Y does not equal -12. THe equation when you plug in different values for X and Y does. In my example, X is 2 and -2 and Y is -4. When you multiply them, you get different answers. Let me know if this is still confusing.

aserghe1 · Answer

(1)
 x^2-y^2<0 
 x^2<y^2 
 |x|<|y| 
Magnitude of y is more than magnitude of x. However we dont know the exact values of x and y.
Not sufficient

(2)
 x^y=\frac{1}{16} 
We can have x=-2 and y=-4, or x=2 and y=-4. Both of these satisfy (2). Notice that they also satisfy (1).

Answer: E

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If x and y are integers, what is the value of xy? (1) x^2 - y^2 < 0

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