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by bblast » Thu May 19, 2011 1:02 am
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-C

answer this as well
[spoiler]if x = 4 the no of divisors / factors x^3 has is 7, correct ?, how can 7 be a multiple of anything else than 1 or 7 ? [/spoiler]
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by pemdas » Thu May 19, 2011 1:32 am
to find divisors/factors we perform factorization of baseline number
st(1) x=4 <> x=2^2 and x^3=2^6 so x^3 has (6+1)=7 factors BUT this is not sufficient as we don't know about y
st(2) y=6 or 2*3 and y^2=2^2 * 3^2, so y^2 has (2+1)(2+1)=9 factors Not sufficient too, as we don't know about x
combined st(1&2): must be sufficient to answer No, as 7 isn't a multiple of 9

re selected/spoiler part @bb you've got two answers Yes or No for 'c'



bblast 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-C

answer this as well
[spoiler]if x = 4 the no of divisors / factors x^3 has is 7, correct ?, how can 7 be a multiple of anything else than 1 or 7 ? [/spoiler]
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by djiddish98 » Thu May 19, 2011 4:51 am
I got tripped up on this one, since I thought A as well.

However, if y = 1, then the number of factors of y^2 = 1, and 7 is a multiple of 1.

Are there any possible values of y^2 that could be a factor of 7 besides 1 - or can the number of divisors in y^2 = 7?

Edit:

I guess 27^2 would have 7 factors? since 27^2 = 3^6?
same with 5^3^2 and so on.
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by GMATGuruNY » Thu May 19, 2011 7:38 am
bblast 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-C

answer this as well
[spoiler]if x = 4 the no of divisors / factors x^3 has is 7, correct ?, how can 7 be a multiple of anything else than 1 or 7 ? [/spoiler]
No math is needed here.

Statement 1: x=4.
Thus, x^3 = 4^3.
No information about y.
Insufficient.

Statement 2: y=6.
Thus, y^2 = 6^2.
No information about x.
Insufficient.

Statements 1 and 2 combined:
Since we know the exact values of x^3 and y^2, we can determine whether the number of factors of the first is a multiple of the number of factors of the second.
Sufficient.

The correct answer is C.

There is no need to determine how many factors x^3 and y^2 each have. We are not being asked to solve the problem.
djiddish98 wrote:I got tripped up on this one, since I thought A as well.

However, if y = 1, then the number of factors of y^2 = 1, and 7 is a multiple of 1.

Are there any possible values of y^2 that could be a factor of 7 besides 1 - or can the number of divisors in y^2 = 7?
If x=4 and y is the cube of a prime number, then x^3 and y^2 will have the same number of factors.
If y=2^3=8, then y^2 = (2^3)^2 = 2^6, which has 6+1 = 7 factors.
If y=3^3=27, then y^2 = (3^3)^2 = 3^6, which has 6+1 = 7 factors.
Since x^3 and y^2 would have the same number of factors, the number of factors of x^3 would be a multiple of the number of factors of y^2.
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by djiddish98 » Thu May 19, 2011 11:18 am
Would it be trickier if they put y = 1 as a statement? Wouldn't that provide the answer without knowing what x was?
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