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:

Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.

The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.

Statement 2 tells us that d > 6, which is a worthless piece of information on its own.

When we combine the 2 statements, we get C.

There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.

Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.

The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.

Statement 2 tells us that d > 6, which is a worthless piece of information on its own.

When we combine the 2 statements, we get C.

There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.

i didnt get how we get "total of seven 5s"..i am able to see only six 5's.

Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.

The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.

Statement 2 tells us that d > 6, which is a worthless piece of information on its own.

When we combine the 2 statements, we get C.

There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.

i didnt get how we get "total of seven 5s"..i am able to see only six 5's.

I think it is C
F = 1*2*3...*30
From A we know 10^d * X = F means..
and F contains for 7 instances of (5*2)
as in... 1*2*3*4...10 has two (5*2 and 10)
11 to 20 has (15 and 20)(meaning another 5 and 2*10 makes 2)
21 to 30 has a 25 and 30 (5 * 5 = 25 and 3 *10...for makes 3 instances of 10)
so d could be from 1 to 7...
from statement 2 u get that d > 6

Thus combining both u get the exact vlue of d...
Hence C

Great to know you are joining Kellogg. A lot was being talked about your last minute interview on Pagalguy (all good though). It was kinda surprise that you got the...