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x and y are two integers greater than 1. Is x^y greater than

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x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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New post 14 Feb 2014, 01:14
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\(x\) and \(y\) are two integers greater than 1. Is \(x^y\) greater than 8?

(1) The sum of ANY two factors of \(x^2\) is even.

(2) The product of ANY two factors of \(y^3\) is odd.

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Re: x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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New post 14 Feb 2014, 01:15
\(x\) and \(y\) are two integers greater than 1. Is \(x^y\) greater than 8?

(1) The sum of ANY two factors of \(x^2\) is even.

From this statement it follows that each factor of \(x^2\) is odd: if even one factor were even, the sum of at least one pair of factors, 1 and that even factor, would be odd. Next, if all the factors of \(x^2\) are odd, then \(x^2\) is odd. For \(x^2\) to be odd, \(x\) must also be odd. The least value of \(x\) is therefore 3 and since \(y\) is greater than 1, then the least value of \(y\) is 2. Thus the least value of \(x^y=3^2=9>8\). Sufficient.

(2) The product of ANY two factors of \(y^3\) is odd.

From this statement it follows that each factor of \(y^3\) is odd: if even one factor were even, the product of at least one pair of factors would be even. Next, if all the factors of \(y^3\) are odd, then \(y^3\) is odd. For \(y^3\) to be odd, \(y\) must also be odd. The least value of \(y\) is therefore 3 and since \(x\) is greater than 1, then the least value of \(x\) is 2. Thus the least value of \(x^y=2^3=8\), but if \(y=odd>3\), then \(x^y>8\). Not sufficient.

Answer: A.
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Re: x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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New post 14 Feb 2014, 03:57
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St1: The sum of ANY two factors of x^2 is even.

This means x is an odd integer. You can try by having x = 3, 5, 7, 9 etc. From stem, y can be 2 at a minimum. If x = 3 and y = 2, x^y = 3^2 = 9 so answer is yes to the stem question. Further, any combination of x and y will always have x^y > 8. Sufficient.

St2: The product of ANY two factors of y^3 is odd.

This means y is an odd integer. You can try by having y = 3, 5, 7, 9 etc. From stem, x can be 2 at a minimum. If x = 2 and y = 3, x^y = 2^3 = 8 so answer is no to the stem question. If x = 3 and y = 3, x^y = 3^3 = 27 so answer is yes to the stem question. Not sufficient.

Answer (C).

Answer (A). I hope I'm right. :|

p.s. I have edited my original answer.
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Re: x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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New post 03 Mar 2014, 21:41
Got A but took 5 minutes to get there!
S1 is basically saying x will be odd.Since x>1 and odd,x=3,5,7,9...
And least y=2
So x^y=9 (least value).Sufficient.

S2 is saying y is odd.This is not sufficient since x=2 or x=3 would give diff answers to the question.
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Re: x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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New post 21 Nov 2017, 11:50
Bunuel wrote:
\(x\) and \(y\) are two integers greater than 1. Is \(x^y\) greater than 8?

(1) The sum of ANY two factors of \(x^2\) is even.


S1) This statement means that x = odd
x = 3
\(x^2\) = 9
Factors of 9 = 1,3,9
Sum of any factor is even.

=> Minimum value of x can be 3.
=> As per the question stem, minimum value of y can be 2.

\(x^y\) = \(3^2\) = 9 > 8

Sufficient.

Quote:
(2) The product of ANY two factors of \(y^3\) is odd.


This statement means that y = odd
y = 3
\(y^3\) = 27
Factors of 27 are 1,3,9,27. Product of any 2 factors is odd.

=> Minimum value of y can be 3.
If x = 2, then
\(x^y\) = \(2^3\) = 8
\(x^y\) = 8

if x = 3, then
\(x^y\) = \(3^3\) = 27
\(x^y\) > 8
Insufficient.

A is the answer.
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Re: x and y are two integers greater than 1. Is x^y greater than &nbs [#permalink] 21 Nov 2017, 11:50
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