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:

Find the power of 80 in 40!??? [#permalink]
18 Oct 2010, 17:17

1

This post received KUDOS

1

This post was BOOKMARKED

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? _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

Re: Find the power of 80 in 40!??? [#permalink]
18 Oct 2010, 20:55

shrouded1 wrote:

Well done Perfect !

Posted from my mobile device

Thanks! man That's a relief _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

Re: Find the power of 80 in 40!??? [#permalink]
19 Oct 2010, 12:42

Expert's post

1

This post was BOOKMARKED

AtifS wrote:

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. _________________

Re: Find the power of 80 in 40!??? [#permalink]
24 Jun 2012, 23:11

AtifS wrote:

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

Re: Find the power of 80 in 40!??? [#permalink]
24 Jun 2012, 23:49

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

Re: Find the power of 80 in 40!??? [#permalink]
31 Jan 2015, 03:27

nadooz wrote:

the step in yellow is unclear for me also

The key thing to see is that 40! / 80^n = Int.

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

Re: Find the power of 80 in 40!??? [#permalink]
16 Feb 2015, 22:14

Since 80 = 16*5 = 2^4*5 (prime factorization)

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

Powers of 5 in 40! = 40/5=(8) - 8/5=(1) 8+1 = 9

Both 16 and 5 are having power of 9 in 40!.

Answer is 9.

gmatclubot

Re: Find the power of 80 in 40!???
[#permalink]
16 Feb 2015, 22:14

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...