Last visit was: 19 Nov 2025, 18:48 It is currently 19 Nov 2025, 18:48
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
Theory|   
User avatar
stne
User avatar
Joined: 27 May 2012
Last visit: 19 Nov 2025
Posts: 1,771
Own Kudos:
1,974
 [5]
Given Kudos: 658
Posts: 1,771
Kudos: 1,974
 [5]
Everything about Factorials on the GMAT 19 Jun 2012, 07:25
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
Bunuel
If you are aiming for 700+ in GMAT you should know 2 important things about factorials:

1. Trailing zeros:
Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

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.

2. Finding the number of powers of a prime number k, in the n!.

What is the power of 3 in 35! ?

The formula is:
\(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

What is the power of 2 in 25!
\(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\)

There is another formula finding powers of non prime in n!, but think it's not needed for GMAT.
Show more

bunuel

How did we decide to use 5 in point no . 1, I mean what is the complete question to the example

Is it no. of trailing zero's when 32! is divided by 5 , In the second case its apparent that we are dividing by the prime factors .

But in order to find the no. of trailing zero's of n! what should we divide by ?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,375
 [1]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,375
 [1]
Everything about Factorials on the GMAT 19 Jun 2012, 07:53
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
stne
Bunuel
If you are aiming for 700+ in GMAT you should know 2 important things about factorials:

1. Trailing zeros:
Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

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.

2. Finding the number of powers of a prime number k, in the n!.

What is the power of 3 in 35! ?

The formula is:
\(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

What is the power of 2 in 25!
\(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\)

There is another formula finding powers of non prime in n!, but think it's not needed for GMAT.

bunuel

How did we decide to use 5 in point no . 1, I mean what is the complete question to the example

Is it no. of trailing zero's when 32! is divided by 5 , In the second case its apparent that we are dividing by the prime factors .

But in order to find the no. of trailing zero's of n! what should we divide by ?
Show more

Not sure understand your question. Anyway:

1st example above asks: "How many zeros are in the end (after which no other digits follow) of 32!?" Here the answer is 32/5+32/5^2=6+1=7. Notice that you take only the quotient into account and that the last denominator (5^2) must be less than numerator.

2nd example above asks: "What is the highest power of 2 in 25!?"

I think it would be easier to understand if you see the questions below:
p-and-q-are-integers-if-p-is-divisible-by-10-q-and-cannot-109038.html
question-about-p-prime-in-to-n-factorial-108086.html
how-many-zeros-does-100-end-with-100599.html
if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html
what-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html
find-the-number-of-trailing-zeros-in-the-product-of-108248.html
find-the-number-of-trailing-zeros-in-the-expansion-of-108249.html
if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html
if-m-is-the-product-of-all-integers-from-1-to-40-inclusive-108971.html
if-10-2-5-2-is-divisible-by-10-n-what-is-the-greatest-106060.html

Hope it helps.
avatar
skamran
avatar
Joined: 04 Sep 2013
Last visit: 06 Aug 2015
Posts: 8
Own Kudos:
1
 [1]
Given Kudos: 5
Posts: 8
Kudos: 1
 [1]
Everything about Factorials on the GMAT 23 Sep 2013, 08:29
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Can someone please explain this part of the above topic??? thanks
"The formula actually counts the number of factors 5 in , 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."
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,375
 [1]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,375
 [1]
Everything about Factorials on the GMAT 24 Sep 2013, 00:35
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
skamran
Can someone please explain this part of the above topic??? thanks
"The formula actually counts the number of factors 5 in , 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."
Show more

Consider 12!. Now, if we prime factorize 12! we'll get that the power of 2 in 12! will naturally be higher than the power of 5. To get one trailing zeros we need one 2 and one 5. Thus if we know what is the power of 5 in 12! we'll know how many trailing zeros 12! has.

Does this make sense?
User avatar
reto
User avatar
Retired Moderator
Joined: 29 Apr 2015
Last visit: 24 Aug 2018
Posts: 716
Own Kudos:
4,292
Given Kudos: 302
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE:Asset Management (Finance: Investment Banking)
Schools: LBS MIF '19
Posts: 716
Kudos: 4,292
Everything about Factorials on the GMAT 26 May 2015, 12:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
noboru



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
Show more

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?

3.04141E+64 =50!
5.31441E+17 =900^6
5.722948210942210000000000E+46 =Quotient (50!/900^6)

Thank you very much
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,375
 [1]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,375
 [1]
Everything about Factorials on the GMAT 27 May 2015, 03:34
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
reto
Bunuel
noboru



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?

3.04141E+64 =50!
5.31441E+17 =900^6
5.722948210942210000000000E+46 =Quotient (50!/900^6)

Thank you very much
Show more

Use better calculator: https://www.wolframalpha.com/input/?i=50%21%2F900%5E6
avatar
bellabear
avatar
Joined: 12 Jan 2017
Last visit: 21 Jun 2018
Posts: 11
Own Kudos:
8
Given Kudos: 392
Location: Thailand
Concentration: Marketing, Operations
Posts: 11
Kudos: 8
Everything about Factorials on the GMAT 19 Dec 2017, 03:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
FACTORIALS

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, Bunuel

--------------------------------------------------------

Definition

The factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\).

For example: \(4!=1*2*3*4=24\).

Properties

  • Factorial of a negative number is undefined.
  • \(0!=1\), zero factorial is defined to equal 1.
  • \(n!=(n-1)!*n\), valid for \(n\geq{1}\).

Trailing zeros:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example: 125,000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^k\leq{n}\)

Example:
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.

Finding the powers of a prime number p, in the n!

The formula is:
\(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}\), where \(k\) must be chosen such that \(p^k\leq{n}\)

Example:
What is the power of 2 in 25!?

\(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\).

_________________________________________________________________________________________________
Questions to practice:
https://gmatclub.com/forum/if-60-is-writ ... 01752.html
https://gmatclub.com/forum/how-many-zero ... 00599.html
https://gmatclub.com/forum/find-the-numb ... 08249.html
https://gmatclub.com/forum/find-the-numb ... 08248.html
https://gmatclub.com/forum/if-n-is-the-p ... 01187.html
https://gmatclub.com/forum/if-m-is-the-p ... 08971.html
https://gmatclub.com/forum/if-p-is-a-nat ... 08251.html
https://gmatclub.com/forum/if-10-2-5-2-i ... 06060.html
https://gmatclub.com/forum/p-and-q-are-i ... 09038.html
https://gmatclub.com/forum/question-abou ... 08086.html
https://gmatclub.com/forum/if-n-is-the-p ... 06289.html
https://gmatclub.com/forum/what-is-the-g ... 05746.html
https://gmatclub.com/forum/if-d-is-a-pos ... 26692.html
https://gmatclub.com/forum/if-10-2-5-2-i ... 06060.html
https://gmatclub.com/forum/how-many-zero ... 42479.html
Show more
Bunuel
could you please explain what this mean ? >> The formula actually counts the number of factors 5 in n!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.
Thank you!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,375
 [3]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,375
 [3]
Everything about Factorials on the GMAT 19 Dec 2017, 03:26
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
gracecud46
Bunuel
FACTORIALS

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, Bunuel

--------------------------------------------------------

Definition

The factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\).

For example: \(4!=1*2*3*4=24\).

Properties

  • Factorial of a negative number is undefined.
  • \(0!=1\), zero factorial is defined to equal 1.
  • \(n!=(n-1)!*n\), valid for \(n\geq{1}\).

Trailing zeros:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example: 125,000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^k\leq{n}\)

Example:
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.

Finding the powers of a prime number p, in the n!

The formula is:
\(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}\), where \(k\) must be chosen such that \(p^k\leq{n}\)

Example:
What is the power of 2 in 25!?

\(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\).

_________________________________________________________________________________________________
Questions to practice:
https://gmatclub.com/forum/if-60-is-writ ... 01752.html
https://gmatclub.com/forum/how-many-zero ... 00599.html
https://gmatclub.com/forum/find-the-numb ... 08249.html
https://gmatclub.com/forum/find-the-numb ... 08248.html
https://gmatclub.com/forum/if-n-is-the-p ... 01187.html
https://gmatclub.com/forum/if-m-is-the-p ... 08971.html
https://gmatclub.com/forum/if-p-is-a-nat ... 08251.html
https://gmatclub.com/forum/if-10-2-5-2-i ... 06060.html
https://gmatclub.com/forum/p-and-q-are-i ... 09038.html
https://gmatclub.com/forum/question-abou ... 08086.html
https://gmatclub.com/forum/if-n-is-the-p ... 06289.html
https://gmatclub.com/forum/what-is-the-g ... 05746.html
https://gmatclub.com/forum/if-d-is-a-pos ... 26692.html
https://gmatclub.com/forum/if-10-2-5-2-i ... 06060.html
https://gmatclub.com/forum/how-many-zero ... 42479.html

Bunuel
Could you please explain what >> The formula actually counts the number of factors 5 in n!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
means from above? Thank you!
Show more

We are getting 0 by multiplying 5 by 2. n! will have more 2's than 5's (in some cases equal number but you cannot have more 5's than 2's). So, if we count the number of 5's we;d be sure that there will be enough 2's for these 5's to make 0. So, the formula counts this number, the number of 5's in n!.

Hope it's clear.
User avatar
targetbusiness
User avatar
Joined: 06 Mar 2017
Last visit: 24 Feb 2021
Posts: 177
Own Kudos:
396
 [1]
Given Kudos: 330
Concentration: Operations, General Management
Schools: Terry '21 (M$)
Schools: Terry '21 (M$)
Posts: 177
Kudos: 396
 [1]
Everything about Factorials on the GMAT 15 May 2018, 03:24
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
aim-high
Hi,

I just went through this thread. I understand point# 1 about trailing zeroes. However, for the life of me, I cant understand Point#2.
Can anyone please explain me what are we really trying to solve in "2. Finding the number of powers of a prime number k, in the n!."

For the question "What is the power of 2 in 25!", the answer is given as 22. What does it mean ?

Thanks
Show more

This is a type of question you see on GMAT at times framed as follows:-
Q. What is the greatest integer k such that 2^k is a factor of 25!?
User avatar
adkikani
User avatar
IIM School Moderator
Joined: 04 Sep 2016
Last visit: 24 Dec 2023
Posts: 1,236
Own Kudos:
1,345
Given Kudos: 1,207
Location: India
WE:Engineering (Other)
Posts: 1,236
Kudos: 1,345
Everything about Factorials on the GMAT 31 Jul 2018, 06:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel chetan2u KarishmaB niks18 GMATPrepNow EMPOWERgmatRichC

What is the rationale behind choosing 5 and its powers in the denominator?
for e.g. in terminating decimals I need to know whether the denominator contains either of 2 or 5 or its powers, else the long division may go indefinitely.

If I do not recall this formula in exam under timed condition, is there any alternate approach to this?
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,001
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,001
 [1]
Everything about Factorials on the GMAT 01 Aug 2018, 05:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
adkikani
Bunuel chetan2u KarishmaB niks18 GMATPrepNow EMPOWERgmatRichC

What is the rationale behind choosing 5 and its powers in the denominator?
for e.g. in terminating decimals I need to know whether the denominator contains either of 2 or 5 or its powers, else the long division may go indefinitely.

If I do not recall this formula in exam under timed condition, is there any alternate approach to this?
Show more

I have explained why you need 2 and/or 5 in the denominator for a terminating decimal in this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/1 ... -the-gmat/

In number properties questions, if you are unable to recall the property, look for a pattern by using some simple numbers.
I am not sure to which question you are referring but if you point it out, I can tell you how to solve it without "knowing" the property.
User avatar
adkikani
User avatar
IIM School Moderator
Joined: 04 Sep 2016
Last visit: 24 Dec 2023
Posts: 1,236
Own Kudos:
1,345
Given Kudos: 1,207
Location: India
WE:Engineering (Other)
Posts: 1,236
Kudos: 1,345
Everything about Factorials on the GMAT 01 Aug 2018, 06:59
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thank you KarishmaB for your two cents.

Below is a relevant quote from your blog:

Quote:
Say, a and b are two integers.

a/b = a * 1/b

For a/b to be terminating, 1/b must be a terminating decimal. What happens when you start dividing 1 by b? You add a decimal point and start adding 0s.You will get 1 followed by as many 0s as you require in the numerator. 10/100/1000/10000 etc have only two prime divisors: 2 and 5. If the denominator has 2s or 5s or both, we will be able to terminate the decimal by choosing the required multiple of 10. If there are any other primes, we will never be able to divide a multiple of 10 completely and hence the decimal will not terminate. It is obvious, isn’t it?
Show more

Why do we not need to need to know whether b is greater than (or not) 1? I could not understand highlighed text in context of long division, Can you elaborate?

I would be grateful if you could explore an additional approach for factorial problem. TIA.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,001
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,001
 [1]
Everything about Factorials on the GMAT 02 Aug 2018, 03:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
adkikani
Thank you KarishmaB for your two cents.

Below is a relevant quote from your blog:

Quote:
Say, a and b are two integers.

a/b = a * 1/b

For a/b to be terminating, 1/b must be a terminating decimal. What happens when you start dividing 1 by b? You add a decimal point and start adding 0s.You will get 1 followed by as many 0s as you require in the numerator. 10/100/1000/10000 etc have only two prime divisors: 2 and 5. If the denominator has 2s or 5s or both, we will be able to terminate the decimal by choosing the required multiple of 10. If there are any other primes, we will never be able to divide a multiple of 10 completely and hence the decimal will not terminate. It is obvious, isn’t it?

Why do we not need to need to know whether b is greater than (or not) 1? I could not understand highlighed text in context of long division, Can you elaborate?

I would be grateful if you could explore an additional approach for factorial problem. TIA.
Show more

If b = 1, a/b is an integer and the question of terminating/non-terminating decimal doesn't arise. b cannot be 0 since then a/b is not defined for b = 0.
b would take values out of 2/3/4/5... etc

When you divide 1 by 3 by long division, how do you do it?
Attachment:
decimals13-05.gif
decimals13-05.gif [ 7.14 KiB | Viewed 12852 times ]
You will keep getting 0s after the decimal but it will never be fully divisible.
Only if you divide by 2 and/or 5 or some combination of them will it be fully divisible. 10/100/1000/10000 etc have only 2 and 5 as factors.
User avatar
ellenckh
User avatar
Joined: 10 Sep 2022
Last visit: 16 Dec 2023
Posts: 19
Own Kudos:
3
Given Kudos: 128
Location: China
Schools: HBS
GPA: 3.92
Schools: HBS
Posts: 19
Kudos: 3
Everything about Factorials on the GMAT 28 Feb 2023, 14:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi, I have a question about the formula for trailing zeros. The example gives us something like 32, which cannot be wholly divided by 5. However, if our number is something like 350, where 5 is a factor of 350, can we still use 5 in the formula? In this case, wouldn't it be that the number of 2 is not equal to the number of 5 any more?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,375
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,375
Everything about Factorials on the GMAT 01 Mar 2023, 01:53
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ellenckh
Hi, I have a question about the formula for trailing zeros. The example gives us something like 32, which cannot be wholly divided by 5. However, if our number is something like 350, where 5 is a factor of 350, can we still use 5 in the formula? In this case, wouldn't it be that the number of 2 is not equal to the number of 5 any more?
Show more

The factorial of a positive integer n, denoted by n!, is composed of the product of integers from 1 to n: n! = 1*2*3*4*5*... Since half of these integers are even, while only one-fifth of the integers are multiples of 5, the power of 2 in n! is always greater than the power of 5 (for n > 1). To calculate the number of trailing zeros in n!, we need to find the power of 5 in prime factorization of n!, since for each 5, there will always be a corresponding 2 in n!, to multiply with it and produce a trailing zero.

For example, the number of trailing zeros in 350! is 86, because the power of 5 in prime prime factorization of n! is 86.

350/5 + 350/25 + 350/125 = 70 + 14 + 2 = 86 (remember to consider only the quotient of the division).

However, the power of 2 in prime prime factorization of n! is 344:

350/2 + 350/4 + 350/8 + 350/16 + 350/32 + 350/64 + 350/128 + 350/256 = 175 + 87 + 43 + 21 + 10 + 5 + 2 + 1 = 344 (remember to consider only the quotient of the division).

Factoring 350! gives:

2^344*3^171*5^86*7^58*11^33*13^28*17^21*19^18*23^15*29^12*31^11*37^9*41^8*43^8*47^7*53^6*59^5*61^5*67^5*71^4*73^4*79^4*83^4*89^3*97^3*101^3*103^3*107^3*109^3*113^3*127^2*131^2*137^2*139^2*149^2*151^2*157^2*163^2*167^2*173^2*179*181*191*193*197*199*211*223*227*229*233*239*241*251*257*263*269*271*277*281*283*293*307*311*313*317*331*337*347*349

Notice that the power of 2 (344) is higher than the power of 5 (86). Additionally, note that the number of trailing zeros in 350! is equal to the power of 5 in its prime factorization.

Hope it helps.
User avatar
ADisHere
User avatar
Joined: 31 Aug 2023
Last visit: 19 Nov 2025
Posts: 127
Own Kudos:
65
Given Kudos: 421
Location: India
Schools: ISB '27 ISB
GMAT Focus 1: 685 Q86 V81 DI82
Schools: ISB '27 ISB
GMAT Focus 1: 685 Q86 V81 DI82
Posts: 127
Kudos: 65
Everything about Factorials on the GMAT 15 May 2024, 11:10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

noboru

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

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
Show more
­Hello Bunnel
here as the factors of 2 are 47 and 5 are 12 so why isn't 10^12 possible? I tried to break it down on my own but I couldn't. Please could you help? 
 
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,375
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,375
Everything about Factorials on the GMAT 15 May 2024, 12:04
Kudos
Add Kudos
Bookmarks
Bookmark this Post
anushridi

Bunuel

noboru

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

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
­Hello Bunnel
here as the factors of 2 are 47 and 5 are 12 so why isn't 10^12 possible? I tried to break it down on my own but I couldn't. Please could you help? 

 
Show more
That's because there is not enough 3's. Please check the highlighted part. 
User avatar
AugmentAdInf
User avatar
Joined: 10 Jan 2023
Last visit: 01 Nov 2025
Posts: 10
Given Kudos: 48
Posts: 10
Kudos: 0
Everything about Factorials on the GMAT 01 Jun 2024, 02:27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
First of all, 'thank you' for summarizing all about factorials in this space. Would it be helpful to add a section on 'Application areas' e.g., achieve efficiency in permutation and combination calculations by simplifying factorials (rather than calculating value).
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,588
Own Kudos:
1,079
Posts: 38,588
Kudos: 1,079
Everything about Factorials on the GMAT 05 Jun 2025, 09:15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

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.
   1   2 
Moderator:
Bunuel
Math Expert
105390 posts