Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 11 Feb 2016, 13:47

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m02 #18

Author Message
Intern
Joined: 29 Dec 2009
Posts: 42
Location: India
Concentration: Finance, Real Estate
Schools: Duke (Fuqua) - Class of 2014
GMAT 1: 770 Q50 V44
GPA: 3.5
WE: General Management (Real Estate)
Followers: 4

Kudos [?]: 13 [0], given: 1

m02 #18 [#permalink]  06 Jun 2010, 03:06
1
This post was
BOOKMARKED
Is the total number of divisors of $$x^3$$ a multiple of the total number of divisors of $$y^2$$?
1. $$x = 4$$
2. $$y = 6$$

OA:
[Reveal] Spoiler:
C

OE:
[Reveal] Spoiler:
To get the divisors of x and y, we need their respective values.

Statement (1) by itself is insufficient. We can only find the divisors of x .

Statement (2) by itself is insufficient. We can only find the divisors of y .

Statements (1) and (2) combined are sufficient. Combining the statements, we have the divisors of x and y .

The way I approach this problem is this:
1. x=4
$$x^3 = 64 = 2^6$$
Total no. of factors = 6+1= 7
Now y is a perfect square. The total no. of factors of a perfect square is always in the form 2k+1.
So, question is: Is 7 a multiple of any number of the form 2k+1?
Answer is: it may be (e.g. when k=3) or may not be (e.g. when k=1, 5,….)
So insufficient

2. y=6
$$y^2 = 36 = 2^2 * 3^2$$
Total no. of factors = (2+1)(2+1) = 9
Now x is a perfect cube. The total no. of factors of a perfect cube is always in the form 3k+1
So, question is: Is a number of the form 3k+1 a multiple of 9
Answer is: No, it can never be.
So, sufficient

And hence, B is my answer.
What’s wrong in this? And doesn’t the OE too simplistic? (In a way, it ignores certain factorization properties of perfect squares and cubes)
Math Expert
Joined: 02 Sep 2009
Posts: 31297
Followers: 5358

Kudos [?]: 62424 [0], given: 9455

Re: m02 #18 [#permalink]  06 Jun 2010, 05:14
Expert's post
deepakdewani wrote:
Is the total number of divisors of $$x^3$$ a multiple of the total number of divisors of $$y^2$$?
1. $$x = 4$$
2. $$y = 6$$

OA:
[Reveal] Spoiler:
C

OE:
[Reveal] Spoiler:
To get the divisors of x and y, we need their respective values.

Statement (1) by itself is insufficient. We can only find the divisors of x .

Statement (2) by itself is insufficient. We can only find the divisors of y .

Statements (1) and (2) combined are sufficient. Combining the statements, we have the divisors of x and y .

The way I approach this problem is this:
1. x=4
$$x^3 = 64 = 2^6$$
Total no. of factors = 6+1= 7
Now y is a perfect square. The total no. of factors of a perfect square is always in the form 2k+1.
So, question is: Is 7 a multiple of any number of the form 2k+1?
Answer is: it may be (e.g. when k=3) or may not be (e.g. when k=1, 5,….)
So insufficient

2. y=6
$$y^2 = 36 = 2^2 * 3^2$$
Total no. of factors = (2+1)(2+1) = 9
Now x is a perfect cube. The total no. of factors of a perfect cube is always in the form 3k+1
So, question is: Is a number of the form 3k+1 a multiple of 9
Answer is: No, it can never be.
So, sufficient

And hence, B is my answer.
What’s wrong in this? And doesn’t the OE too simplistic? (In a way, it ignores certain factorization properties of perfect squares and cubes)

When solving this question I made the same assumption as you did and arrived to answer B with same logic as you did. But answer B is not correct. The trick here is that we are not told that $$x$$ and $$y$$ are integers, so for (2) the logic would be correct ONLY for integers. But if $$y^2=36=x^3$$ then the total number of divisors of $$x^3$$ will obviously be equal to the total number of divisors of $$y^2$$ (note that in this case $$x=\sqrt[3]{36}\neq{integer}$$). Hence insufficient.

Though I must say that this is not GMAT type of question, as every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So I would suggest to change the question as follows:

If $$x$$ and $$y$$ are positive integers, is the total number of divisors of $$x^3$$ a multiple of the total number of divisors of $$y^2$$?

(1) $$x = 4$$
(2) $$y = 6$$

In this case: A. the question will meet the GMAT standards and B. the question will be 750+ difficulty level, with elegant solution.

Hope it helps.
_________________
Intern
Joined: 29 Dec 2009
Posts: 42
Location: India
Concentration: Finance, Real Estate
Schools: Duke (Fuqua) - Class of 2014
GMAT 1: 770 Q50 V44
GPA: 3.5
WE: General Management (Real Estate)
Followers: 4

Kudos [?]: 13 [0], given: 1

Re: m02 #18 [#permalink]  06 Jun 2010, 06:35
Quote:
The trick here is that we are not told that $$x$$ and $$y$$ are integers

Yes, that's would I could also think of.

Quote:
Though I must say that this is not GMAT type of question, as every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

Agree -without x & y being integers, this question is a bit absurd.

Thanks a bunch.
Intern
Joined: 24 Apr 2011
Posts: 25
Followers: 0

Kudos [?]: 2 [0], given: 5

Re: m02 #18 [#permalink]  04 Jun 2011, 13:25
is there any way to simplify the question stem to what we are looking for?
Re: m02 #18   [#permalink] 04 Jun 2011, 13:25
Similar topics Replies Last post
Similar
Topics:
1 M02 Q18 Is the total number of divisors of x^3 a multiple 3 18 Jan 2013, 10:07
36 M02 #19 26 26 Dec 2008, 17:28
15 m02 #21 28 14 Nov 2008, 12:33
40 m02#24 14 05 Nov 2008, 07:21
22 m02#8 16 03 Nov 2008, 15:24
Display posts from previous: Sort by

# m02 #18

Moderators: Bunuel, WoundedTiger

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.