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integers

by cans » Wed Jun 08, 2011 2:58 am
What is the value of (x-y)^4?

1. The product of x and y is 7

2. x and y are integers
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by krishnasty » Wed Jun 08, 2011 3:25 am
cans wrote:What is the value of (x-y)^4?

1. The product of x and y is 7

2. x and y are integers
(x-y)^4 = x^4 + y^4 + 2x^2y^2

from 1) xy = 7. no idea of x and y independently. Insufficient
from 2) no idea of x and y. Insufficent

combining both, the only way 2 integerts would have product as 7 would be 1 and 7 ( or -1 and -7)
hence, x = 1 and y = 7 ( or vice versa)
Hence, (x-y)^4 can be found out.

IMO, C
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by Shankenstein » Wed Jun 08, 2011 3:38 am
In 1)
For the product to be 7 it has to be 7 and 1 or not an integer.

In 2) It says X and Y are integers which is ...-3,-2,-1,0,1,2,3....

If we combine both we get X=7, Y=1 or X=1 and Y=7 (or their negative counterparts)so it could be 6^4 or (-6)^4. in anyway it would be same.
Therefor answer is C.
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by bubbliiiiiiii » Wed Jun 08, 2011 3:44 am
What is the value of (x-y)^4?

What we have to find? - the consistent value of (x-y)^4.

1. The product of x and y is 7
X and Y could be, (1,7), (3.5, 2) ... and x-y could vary accordingly.
INSUFFICIENT

2. x and y are integers
X and Y could be, (1,7), (5,1), (10,1) ... and x-y could vary accordingly.
INSUFFICIENT

Since A and B are neither sufficient we are left with C and E.

Combining i and ii,

we get (x,y) to be (1,7) or (7,1) => (x-y) is either -6 or 6.

Since, (x-y) is raised to power of 4, which is even, the number is always positive irrespective of its sign.

Thus, we can conclude a consistent value of (x-y)^4.

Answer (C)
Regards,

Pranay
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