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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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\(x\) 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.
[Reveal] Spoiler: OA

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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.

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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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 15 Feb 2014, 06:44
\(x\) 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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, 22:41
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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Re: x and y are two integers greater than 1. Is x^y greater than   [#permalink] 11 Aug 2017, 19:50
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