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:

This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of non-prime number in \(n!\) we first do prime-factorization of the non-prime number and then find the powers of each prime number in \(n!\) one by one using the following formula \(\frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}}\) such that \(p^x <n\), where \(p\) is the prime number.

Let's suppose, we want to find the powers of \(80\) in \(40!\). Prime factorization of \(80=2^4 * 5^1\). Now first find the power of \(2\) in \(40!\); \(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\) Now find the powers of \(5\) in \(40!\); \(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\)

And \(40!=80^x*q=(2^4 * 5^1)^x*q\), where \(q\) is the quotient and \(x\) is any power of \(80\), now from above calculation \(40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q\), So we have \(80\) in the power of \(9\) in \(40!\).

Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values (\(80\) & \(40!\)) randomly and tried to apply the rule/method and I was confused whether I am (or should it be am I? ) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh?
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of non-prime number in \(n!\) we first do prime-factorization of the non-prime number and then find the powers of each prime number in \(n!\) one by one using the following formula \(\frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}}\) such that \(p^x <n\), where \(p\) is the prime number.

Let's suppose, we want to find the powers of \(80\) in \(40!\). Prime factorization of \(80=2^4 * 5^1\). Now first find the power of \(2\) in \(40!\); \(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\) Now find the powers of \(5\) in \(40!\); \(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\)

And \(40!=80^x*q=(2^4 * 5^1)^x*q\), where \(q\) is the quotient and \(x\) is any power of \(80\), now from above calculation \(40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q\), So we have \(80\) in the power of \(9\) in \(40!\).

Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values (\(80\) & \(40!\)) randomly and tried to apply the rule/method and I was confused whether I am (or should it be am I? ) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh?

wanted to ask whether questions like these have been asked

You'll need only to know how to determine the number of trailing zeros and the power of primes in n! (everything-about-factorials-on-the-gmat-85592.html), the above example is out of the scope of GMAT.
_________________

This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of non-prime number in \(n!\) we first do prime-factorization of the non-prime number and then find the powers of each prime number in \(n!\) one by one using the following formula \(\frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}}\) such that \(p^x <n\), where \(p\) is the prime number.

Let's suppose, we want to find the powers of \(80\) in \(40!\). Prime factorization of \(80=2^4 * 5^1\). Now first find the power of \(2\) in \(40!\); \(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\) Now find the powers of \(5\) in \(40!\); \(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\)

And \(40!=80^x*q=(2^4 * 5^1)^x*q\), where \(q\) is the quotient and \(x\) is any power of \(80\), now from above calculation \(40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q\), So we have \(80\) in the power of \(9\) in \(40!\). Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values (\(80\) & \(40!\)) randomly and tried to apply the rule/method and I was confused whether I am (or should it be am I? ) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh?

Can u explain more the step in yellow, please? Thanks
_________________

I hated every minute of training, but I said: "Don't quit. Suffer now and live the rest of your life as a champion." Muhammad Ali

Hi Anan, Not to worry much as this particular concept has been explained very well was Bunuel in math book as well as in one of the topics related to factorial.

This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of non-prime number in \(n!\) we first do prime-factorization of the non-prime number and then find the powers of each prime number in \(n!\) one by one using the following formula \(\frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}}\) such that \(p^x <n\), where \(p\) is the prime number.

Let's suppose, we want to find the powers of \(80\) in \(40!\). Prime factorization of \(80=2^4 * 5^1\). Now first find the power of \(2\) in \(40!\); \(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\) Now find the powers of \(5\) in \(40!\); \(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\)

And \(40!=80^x*q=(2^4 * 5^1)^x*q\), where \(q\) is the quotient and \(x\) is any power of \(80\), now from above calculation \(40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q\), So we have \(80\) in the power of \(9\) in \(40!\). Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values (\(80\) & \(40!\)) randomly and tried to apply the rule/method and I was confused whether I am (or should it be am I? ) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh?

Can u explain more the step in yellow, please? Thanks

After factoring and finding the respective powers for 5 and 2 ... manipulate the original equation in the question stem to 40! = 80^n * int.

So, we know that 40! will equal 80^n * some int. value (in this case, we'll mark that int. as "p" -- for simplicity's sake)

Since 80^n = (5^9 * 2^38) --> We now have 40! = (5^9 * 2^38) * p

Now, we need to match the higher power down to the lower power. We need to deduce 2^38 to the 9th power, so that it'll match up with 5^9. Knowing that when we factorized 80^n ... we were left with 5^n and 2^4n ...

Using 2^4 --> we want to reduce the 38 down by dividing 4 into 38. Well, that obviously won't work. So, we take out 2^2:

40! = (5^9 * 2^36) * 2^2 * p

Then, since we have 2^4 as a factor of 80, we plug that in for 2^36. We divide 4 into 36, and left with this ---> 40! = (5^9 * (2^4)^9) * 2^2 * int

2^4 becomes 8: 40! = (5^9 * 16^9) * 2^2 * int -----> 40! = (5 * 16)^9 * 2^2 * int ----> 40! = (80)^9 * 2^2 * int

40/2= (20) - 20/2=(10) - 10/2=(5) - 5/2=(2) - 2/2=(1) (only quotient without remainder) Total powers of 2 in 40! = 20+10+5+2+1 = 38 So power of 2^4 in 4! = 38/4 = 9

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.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...