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divisible by 3

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divisible by 3

by pappueshwar » Tue Mar 13, 2012 7:39 am
If x and y are positive integers, is x^4 -y^4 divisible by 3?

1) x-y is divisible by 3.
2) x+y is divisible by 3.


OA IS D.

Each stmnt is enough to solve the probelm becoz once we solve the problem taking numbers, the 3 in the denominator gets cancelled ?
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by killer1387 » Tue Mar 13, 2012 8:04 am
pappueshwar wrote:If x and y are positive integers, is x^4 -y^4 divisible by 3?

1) x-y is divisible by 3.
2) x+y is divisible by 3.


OA IS D.

Each stmnt is enough to solve the probelm becoz once we solve the problem taking numbers, the 3 in the denominator gets cancelled ?
x^4 -y^4 is divisible by x-y and x+y

1) sufficient
2)sufficient
hence D

HTH
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by tpr-becky » Tue Mar 13, 2012 11:52 am
to answer your first question - if you pick number and the three always cancels out then yes, the statement is sufficient. However, you can also look at this problem algebraically. For something to be divisible that means that the denominator divides evenly into the numerator, thus

(x^4 - y ^4)/3 will be an integer.

noticing the even exponents that are the same you may be able to recognize the common quadratic form and see that (x^4 - y^4) = (x^2 + y^2)(x^2 - y^2) - you can then factor out the second term one more time to get (x64 - y^4) = (x^2 + y^2) (x + y ) (x - y) - this is what we need to check for divisibilty by three.

Statement 1 says x - y is divisible by three - that means that the term in the numerator will cancel out the three and thus the entire equation is divisible by three. Sufficient.

Statement 2 says x + y is divisible by three - this term is also in the numerator and will thus cancel out the three in the denonminator - Sufficient.

Since both are sufficient, the answer is D.
Becky
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The Princeton Review
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