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:

Hey gmatblackbelt, Im not following the logic above for statement 2.

We know that A^6 = (2^7*3^2*5)k ... where do we go from here ?

A^6 does not equal 2^7*3^2*5*7

8! is divisble by A^6 the only way that this could be is for A to equal 2. It cannot equal 1 since the main stem said A and N are bothg greater than 1.

(2) \(n=6\) --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence \(a=2\). Sufficient.

Answer: B.

Hi bunuel, Thank you for your instruction. I have another query: how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?
_________________

Consider giving me kudos if you find my explanations helpful so i can learn how to express ideas to people more understandable.

(2) \(n=6\) --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence \(a=2\). Sufficient.

Answer: B.

Hi bunuel, Thank you for your instruction. I have another query: how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?

8! has 2^7 as its factor. Thus maximum exponent of 4 is 3; \((2^2)^3*2=(4)^3*2\)

Hi bunuel, Thank you for your instruction. I have another query: how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?

8! has 2^7 as its factor. Thus maximum exponent of 4 is 3; \((2^2)^3*2=(4)^3*2\)

When looking at Statement 1, we know a^n = 64 and a and n are positive integers greater than 1, so a could be 2, 4 or 8 (since 2^6 = 4^3 = 8^2 = 64). The information in the stem isn't actually important here.

When we look at Statement 2, we know that 8! is divisible by a^6. If we prime factorize 8!, we find:

We need this prime factorization to be divisible by a^6 where a > 1. Looking at the prime factorization, the only possibility is that a = 2 (since for any other prime p besides 2, we can't divide the factorization above by p^6, nor can we divide the factorization above by (2^2)^6 = 2^12).
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: If the integers a and n are greater than 1 and the product [#permalink]

Show Tags

30 Jan 2015, 14:57

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Military MBA Acceptance Rate Analysis Transitioning from the military to MBA is a fairly popular path to follow. A little over 4% of MBA applications come from military veterans...

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...