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m02 #18

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m02 #18 [#permalink] New post 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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Re: m02 #18 [#permalink] New post 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 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.

Answer: C.

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

Answer: B.

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.
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Intern
Intern
User avatar
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 [?]: 10 [0], given: 1

Re: m02 #18 [#permalink] New post 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.
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Re: m02 #18 [#permalink] New post 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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