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 350,000 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:

S is composed of 2 prime integers.... q^3 cannot give a prime integer (ex 11^3, 13^3 are not prime integers), so it must be equal to 1 to not interfer and to make it possible for S to be a multiple of 2 primes integers, provided by p here.

This is abad question though. When combining the 2 statements we need to assume both are true. So by stmt 1 we are told that S is a product of 2 prime ints > 10. Using stmt2, q^3 cannot exist at all

This is abad question though. When combining the 2 statements we need to assume both are true. So by stmt 1 we are told that S is a product of 2 prime ints > 10. Using stmt2, q^3 cannot exist at all

ST. 1 says S is the product of two prime numbers which are both >10
ST. 2 gives us S= P*Q^3

which means P is a prime>10
and Q^3 is a prime>10

hence Q^3 needs to be a perfect cube + a prime> 10
perfect cubes >10 are 27, 64, 125, none of which are primes
there is no perfect cube which is also a prime> 10
how can we take Q to be equal to 1, that would make Q^3= 1 and 1<10

S is composed of 2 prime integers.... q^3 cannot give a prime integer (ex 11^3, 13^3 are not prime integers), so it must be equal to 1 to not interfer and to make it possible for S to be a multiple of 2 primes integers, provided by p here.

Ah ok now I see. I thought S1 was sayin that S is the product of 2 primes that are both greater than 10.

S is composed of 2 prime integers.... q^3 cannot give a prime integer (ex 11^3, 13^3 are not prime integers), so it must be equal to 1 to not interfer and to make it possible for S to be a multiple of 2 primes integers, provided by p here.

Ah ok now I see. I thought S1 was sayin that S is the product of 2 primes that are both greater than 10.

Was like how can this be C?

exactly, I am not sure why everyone is agreeing that Q is 1 when ST. 1& 2 implies clearly that Q^3 must be a prime> 10.
1 is not a prime and it is not greater than 10.