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Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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10 00:00

Difficulty:   95% (hard)

Question Stats: 41% (01:52) correct 59% (02:16) wrong based on 140 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.
Math Expert V
Joined: 02 Sep 2009
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Official Solution:

(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.

Intern  Joined: 27 Aug 2018
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Hi Bunuel,

Could you please help me understand this part....."From this statement it follows that each factor of X square 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."

Since even + even = even AND odd + odd = even, why cannot we assume that all factors of X are even OR all factors of X are odd. In the first case lowest X would be 2(4)=8 and lowest Y is 2, therefore 8^2 > 8. Answer to question is Yes. In the 2nd case, lowest X would be 1(3)=3 and lowest Y = 2, therefore 3^2 >8. Answer to question is Yes. Sufficient.

Is my working wrong?
Math Expert V
Joined: 02 Sep 2009
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achatterjee02 wrote:
Hi Bunuel,

Could you please help me understand this part....."From this statement it follows that each factor of X square 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."

Since even + even = even AND odd + odd = even, why cannot we assume that all factors of X are even OR all factors of X are odd. In the first case lowest X would be 2(4)=8 and lowest Y is 2, therefore 8^2 > 8. Answer to question is Yes. In the 2nd case, lowest X would be 1(3)=3 and lowest Y = 2, therefore 3^2 >8. Answer to question is Yes. Sufficient.

Is my working wrong?

An integer cannot have all its factors even because 1 is a factor of every integer so at least one factor of very integer is odd.
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(1) The sum of ANY two factors of $$x^2$$ is even.

Let x be 3. So $$x^2 = 9$$. Sum of any two factors of 9 is even. The minimum value x can take is 3. As y>1, $$x^y = 3^y$$ and the minimum value of $$x^y$$ is 9 which is greater than 8.
SUFFICIENT.

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

Let y=3, $$y^3 = 27$$. The product of any two factors of y is odd.
$$x^y = x^3$$
Now if x = 2, then $$x^y = 8$$. The answer to the question is NO
If x = 3, $$x^y > 8$$. The answer to the question is YES.
INSUFFICIENT.

OPTION: A
Intern  B
Joined: 23 Dec 2015
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Hi Bunuel,

if x=2, then x^2 = 4, factors are 1,2 4..then sum of any 2 factors is even . .2+4 - 6 even? and since y>1, y=2, so 2^2 = 4<8 makes stmt 1 insufficient? No. Please reply.
Math Expert V
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san01sin wrote:
Hi Bunuel,

if x=2, then x^2 = 4, factors are 1,2 4..then sum of any 2 factors is even . .2+4 - 6 even? and since y>1, y=2, so 2^2 = 4<8 makes stmt 1 insufficient? No. Please reply.

(1) says that the sum of ANY two factors of x^2 is even.

How is that true for 4? 1 + 2 = 3 = odd. Re: M30-05   [#permalink] 12 Aug 2019, 22:32
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# M30-05

Moderators: chetan2u, Bunuel  