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# m02 #18

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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 [?]: 15 [0], given: 1

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06 Jun 2010, 03:06
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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 correct answer is C.

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)
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Joined: 02 Sep 2009
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Re: m02 #18 [#permalink]

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06 Jun 2010, 05:14
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 correct answer is C.

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 [?]: 15 [0], given: 1

Re: m02 #18 [#permalink]

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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
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Re: m02 #18 [#permalink]

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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
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# m02 #18

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