GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 05 Dec 2019, 09:57

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M30-05

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

16 Sep 2014, 01:45
10
00:00

Difficulty:

95% (hard)

Question Stats:

41% (01:52) correct 59% (02:16) wrong based on 140 sessions

### HideShow timer Statistics

$$x$$ 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
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

16 Sep 2014, 01:45
1
3
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
Posts: 1

### Show Tags

19 Oct 2018, 22:50
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
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

20 Oct 2018, 02:02
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.
Senior Manager
Joined: 13 Jan 2018
Posts: 341
Location: India
Concentration: Operations, General Management
GMAT 1: 580 Q47 V23
GMAT 2: 640 Q49 V27
GPA: 4
WE: Consulting (Consulting)

### Show Tags

03 Jan 2019, 20:53
(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
Joined: 23 Dec 2015
Posts: 3

### Show Tags

12 Aug 2019, 19:47
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
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

12 Aug 2019, 22:32
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
Display posts from previous: Sort by

# M30-05

Moderators: chetan2u, Bunuel