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Math Expert V
Joined: 02 Sep 2009
Posts: 55618
x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 45% (02:39) correct 55% (02:22) wrong based on 203 sessions

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 55618
Re: 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.

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.

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Manager  Joined: 04 Jan 2014
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GMAT 1: 660 Q48 V32 GMAT 2: 630 Q48 V28 GMAT 3: 680 Q48 V35 Re: x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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1
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 (A). I hope I'm right. p.s. I have edited my original answer.
Manager  Joined: 20 Dec 2013
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Re: x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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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.
Senior Manager  G
Joined: 06 Jul 2016
Posts: 362
Location: Singapore
Concentration: Strategy, Finance
Re: x and y are two integers greater than 1. Is x^y greater than  [#permalink]

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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.

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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] 29 Jan 2019, 04:00
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