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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Total number of divisors for a perfect cube \(x^3\) is one more than a multiple of 3 i.e \(3m+1\) divisors Total number of divisors for a perfect square \(y^2\) will be odd i.e \(2n+1\) divisors

1. \(x=4\)

\(4^3\) has 7 divisors if you check. 7 is not a multiple of anything except 7 and 1. So \(y^2\) can have 7 divisors if say \(y=64\), or \(y^2\) can have 1 divisor, where \(y=1\). NOT SUFF.

2. \(y=6\) \(Y^2 = 36\) has 9 total divisors. \(x^3\) has \(3m+1\) total divisors, which cannot be divided by 9. SUFF.

Total number of divisors for a perfect cube \(x^3\) is one more than a multiple of 3 i.e \(3m+1\) divisors Total number of divisors for a perfect square \(y^2\) will be odd i.e \(2n+1\) divisors

1. \(x=4\)

\(4^3\) has 7 divisors if you check. 7 is not a multiple of anything except 7 and 1. So \(y^2\) can have 7 divisors if say \(y=64\), or \(y^2\) can have 1 divisor, where \(y=1\). NOT SUFF.

2. \(y=6\) \(Y^2 = 36\) has 9 total divisors. \(x^3\) has \(3m+1\) total divisors, which cannot be divided by 9. SUFF.

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. y^2=36=x^3 then total number of divisors of x^3 is equal to the total number of divisors of y^2. Hence insufficient.

Answer: C.

amitgovin wrote:

my question: doesn't a perfect square always have an odd number of factors?

Tips about the perfect square:

1. The number of distinct factors of a perfect square is ALWAYS ODD. 2. The sum of distinct factors of a perfect square is ALWAYS ODD. 4. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. 5. Perfect square always has even number of powers of prime factors.
_________________

Thanks for pointing out the catch, Bunuel. If we can't assume they are integers, then none of the logic really holds true. How do you define a divisor for 1.53 - is it fair to say that the quotient must be an integer?

So would it then be right to skip straight to C since we haven't been told they are integers?

Thanks for pointing out the catch, Bunuel. If we can't assume they are integers, then none of the logic really holds true. How do you define a divisor for 1.53 - is it fair to say that the quotient must be an integer?

So would it then be right to skip straight to C since we haven't been told they are integers?

Well you are partially right. Remember we are not asked to determine # of divisors of x or y, which may not be the integers, but the divisors of x^3 and y^2. x can be 3^1/3 which is not an integer and it makes no sense to ask about the factors of x. But x^3=3 and 3 has 2 distinct factors. So this is not the problem in the question.

BUT you are right about the factors of non integers and here is the tip from GMAT tutor Ian Stewart:

Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So we can say that this question is not real GMAT one.
_________________

B - 2 is sufficient Total number of divisors for a perfect cube \(x^3\) is one more than a multiple of 3 i.e \(3m+1\) divisors Total number of divisors for a perfect square \(y^2\) will be odd i.e \(2n+1\) divisors

Hi.

whats m & n in your statements above?

TX
_________________

i love kudos consider giving them if you like my post!!

3m+1 is to show that the number is one more than a multiple of 3. And 2n + 1 is to show that the number is odd. We do not know exactly what m, n as it will depend on each value for x^3 and y^2.

In general, m,n are integers from {0,1,2,3,4,5,..}