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:

Re: Everything about Factorials on the GMAT [#permalink]

Show Tags

11 Jun 2014, 13:18

How many zeros are in the end (after which no other digits follow) of 32!?

\frac{32}{5}+\frac{32}{5^2}=6+1=7. Notice that the denominators must be less than or equal to 32 also notice that we take into account only the quotient of division (that is \frac{32}{5}=6 not 6.4). Therefore, 32! has 7 trailing zeros.

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero. how come we divide 32 by 5?

Re: Everything about Factorials on the GMAT [#permalink]

Show Tags

11 Jun 2014, 14:13

Expert's post

sagnik2422 wrote:

How many zeros are in the end (after which no other digits follow) of 32!?

\frac{32}{5}+\frac{32}{5^2}=6+1=7. Notice that the denominators must be less than or equal to 32 also notice that we take into account only the quotient of division (that is \frac{32}{5}=6 not 6.4). Therefore, 32! has 7 trailing zeros.

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero. how come we divide 32 by 5?

Well, this is done according to the formula given there. _________________

Re: Everything about Factorials on the GMAT [#permalink]

Show Tags

22 Apr 2015, 01:07

A more generalised method for finding powers of non prime numbers in a factorial:

Assuming we have to find the power of 12 in 30!

Now, 30 = 2 * 3 * 5.

Find what the highest power of each prime is in the given factorial (by the same method as given by Bunuel)

We have 25 + 12 + 6 + 3 + 1 = 47 2s in 50!

16 + 5 + 1 = 22 3s in 50!

10 + 2 = 12 5s in 50!

We need a combination of one 2, one 3 and one 5 to get 30. In the given factorial we have only 12 5s. Hence only twelve 30s are possible. So the greatest power of 30 in 50! is 12.

Similarly we can find powers of 4 (number of pairs of 2) etc.

Re: Everything about Factorials on the GMAT [#permalink]

Show Tags

26 May 2015, 12:51

Bunuel wrote:

noboru wrote:

It's better to illustrate it on the example: How many powers of 900 are in 50! \(900=2^2*3^2*5^2\)

Find the power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\)

= \(2^{47}\)

Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)

=\(3^{22}\)

Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\)

=\(5^{12}\)

We need all of them (2,3,5) to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50! 900^6

If I go to excel and calculate factorial 50! and then divide by 900^6, i still have room for 900^7 which then still yields a positive result. Could you explain that further?

Re: Everything about Factorials on the GMAT [#permalink]

Show Tags

27 May 2015, 03:34

1

This post received KUDOS

Expert's post

reto wrote:

Bunuel wrote:

noboru wrote:

It's better to illustrate it on the example: How many powers of 900 are in 50! \(900=2^2*3^2*5^2\)

Find the power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\)

= \(2^{47}\)

Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)

=\(3^{22}\)

Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\)

=\(5^{12}\)

We need all of them (2,3,5) to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50! 900^6

If I go to excel and calculate factorial 50! and then divide by 900^6, i still have room for 900^7 which then still yields a positive result. Could you explain that further?

Re: Everything about Factorials on the GMAT [#permalink]

Show Tags

22 Sep 2015, 15:58

Bunuel wrote:

noboru wrote:

Point 1 is just point 2 for k=5, isnt it?

Yes, it is. I've separated them as for GMAT generally we need only trailing zeros and almost never other prime's power.

shalva wrote:

Can you please post this one too? It's still interesting, though may not be usable for GMAT.

It's better to illustrate it on the example: How many powers of 900 are in 50! \(900=2^2*3^2*5^2\)

Find the power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\)

= \(2^{47}\)

Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)

=\(3^{22}\)

Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\)

=\(5^{12}\)

We need all of them (2,3,5) to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50! 900^6

Can someone please explain why 900 is there 6 times versus 12 ?

there are 12 5's in 50!

30 = 2 * 3 * 5 30 x 30 = 900 = 2^2 x 3^2 x 5^2

I understand that 900 is a pair of the primes 2 , 3 and 5 but if there are a total of 12 5's then 6 would be in the first 30 and 6 in the next 30 . 6 + 6 = 12 so 900 ^ 12 would seem more correct?

Re: Everything about Factorials on the GMAT [#permalink]

Show Tags

04 Nov 2015, 04:16

Expert's post

ayushkhatri wrote:

Bunuel wrote:

noboru wrote:

Point 1 is just point 2 for k=5, isnt it?

Yes, it is. I've separated them as for GMAT generally we need only trailing zeros and almost never other prime's power.

shalva wrote:

Can you please post this one too? It's still interesting, though may not be usable for GMAT.

It's better to illustrate it on the example: How many powers of 900 are in 50! \(900=2^2*3^2*5^2\)

Find the power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\)

= \(2^{47}\)

Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)

=\(3^{22}\)

Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\)

=\(5^{12}\)

We need all of them (2,3,5) to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50! 900^6

Can someone please explain why 900 is there 6 times versus 12 ?

there are 12 5's in 50!

30 = 2 * 3 * 5 30 x 30 = 900 = 2^2 x 3^2 x 5^2

I understand that 900 is a pair of the primes 2 , 3 and 5 but if there are a total of 12 5's then 6 would be in the first 30 and 6 in the next 30 . 6 + 6 = 12 so 900 ^ 12 would seem more correct?

ayushkhatri In the prime factorization of 900 , the highest power of 5 is 2 . 900=\(2^{2}\)∗\(3^{2}\)∗\(5^{2}\)

as per the calculations in the previous posts , we know power of 5 in 50! is 12 . So the 12 powers of 5 can be divided among 6 pairs . We divide in pairs because 900 has 5^2 .

Just think for a moment , if it was indeed 900 ^ {12} , then we will have (5^2)^12 that is 5^24 . There will be 12 powers of 30 in 50! . _________________

When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford +1 Kudos if you find this post helpful

gmatclubot

Re: Everything about Factorials on the GMAT
[#permalink]
04 Nov 2015, 04:16

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...