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:

C. n=14
is it m that the qs was referring to? be careful with your typing, it freaked me out trying to find P

m = 1x2x3x4x...x30
there are 10 multiples of 3 (3x1, 3x2, 3x3,...3x10)
3 mutilples of 3^2 (9x1, 9x2, 9x3)
1 multiple of 3^3 (27)
therefore the number of times factor 3 appears in the product m is 10+3+1 = 14.
_________________

Carson, this is what I tell my students about any problem that has factors, multiples, products, divisors, or any other reference to perfect integers:

The number in the denominator must disappear completely to make an integer.

If I know that x/y is an integer, than I know that somehow, y will cancel with x.

This problem is broken down essentially to 30!/3^n, and we know it's an integer, because 3^n is a factor of 30!. Since we're just talking about a string of 3's on the bottom, we need to know how many 3's are on top to figure out the maximum number on the bottom. And since the 3's are coming only from the multiples of 3 in 30!, we need only to consider, as jpv and the others said, 3,6,9,13,15,18,21,24,27,30. Prime factor each of these numbers, and you'll find that there are 14 3's among them.

So if n was 15, let's say, then there would be 15 3's on the bottom and only 14 on top, and the bottom wouldn't canel out completely, and you wouldn't have an integer. So it has to be 14.