Aug 18 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes. Aug 19 08:00 AM PDT  09:00 AM PDT Join a 4day FREE online boot camp to kick off your GMAT preparation and get you into your dream bschool in R1.**Limited for the first 99 registrants. Register today! Aug 20 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Aug 20 09:00 PM PDT  10:00 PM PDT Take 20% off the plan of your choice, now through midnight on Tuesday, 8/20 Aug 22 09:00 PM PDT  10:00 PM PDT What you'll gain: Strategies and techniques for approaching featured GMAT topics, and much more. Thursday, August 22nd at 9 PM EDT
Author 
Message 
TAGS:

Hide Tags

Director
Joined: 27 May 2012
Posts: 835

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
19 Jun 2012, 07:25
Bunuel wrote: 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 nonnegative 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 ?
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 57022

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
19 Jun 2012, 07:53
stne wrote: Bunuel wrote: 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 nonnegative 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 ? 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: pandqareintegersifpisdivisibleby10qandcannot109038.htmlquestionaboutpprimeintonfactorial108086.htmlhowmanyzerosdoes100endwith100599.htmlifnistheproductofintegersfrom1to20inclusive106289.htmlwhatisthegreatestvalueofmsuchthat4misafactorof105746.htmlfindthenumberoftrailingzerosintheproductof108248.htmlfindthenumberoftrailingzerosintheexpansionof108249.htmlifdisapositiveintegerandfistheproductofthefirst126692.htmlifmistheproductofallintegersfrom1to40inclusive108971.htmlif10252isdivisibleby10nwhatisthegreatest106060.htmlHope it helps.
_________________



Intern
Joined: 04 Sep 2013
Posts: 12

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
23 Sep 2013, 08:29
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."



Math Expert
Joined: 02 Sep 2009
Posts: 57022

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
24 Sep 2013, 00:35
skamran wrote: 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." 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?
_________________



Retired Moderator
Joined: 29 Apr 2015
Posts: 827
Location: Switzerland
Concentration: Economics, Finance
WE: Asset Management (Investment Banking)

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? 3.04141E+64 =50! 5.31441E+17 =900^6 5.722948210942210000000000E+46 =Quotient (50!/900^6) Thank you very much
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!
PS Please send me PM if I do not respond to your question within 24 hours.



Math Expert
Joined: 02 Sep 2009
Posts: 57022

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
27 May 2015, 03:34
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? 3.04141E+64 =50! 5.31441E+17 =900^6 5.722948210942210000000000E+46 =Quotient (50!/900^6) Thank you very much Use better calculator: http://www.wolframalpha.com/input/?i=50%21%2F900%5E6
_________________



Intern
Joined: 12 Jan 2017
Posts: 12
Location: Thailand
Concentration: Marketing, Operations

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
19 Dec 2017, 03:24
Bunuel wrote: FACTORIALSThis post is a part of [ GMAT MATH BOOK] created by: Bunueledited by: bb, Bunuel DefinitionThe factorial of a nonnegative 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.
 \(n!=(n1)!*n\), 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 nonnegative 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: http://gmatclub.com/forum/if60iswrit ... 01752.htmlhttp://gmatclub.com/forum/howmanyzero ... 00599.htmlhttp://gmatclub.com/forum/findthenumb ... 08249.htmlhttp://gmatclub.com/forum/findthenumb ... 08248.htmlhttp://gmatclub.com/forum/ifnisthep ... 01187.htmlhttp://gmatclub.com/forum/ifmisthep ... 08971.htmlhttp://gmatclub.com/forum/ifpisanat ... 08251.htmlhttp://gmatclub.com/forum/if10252i ... 06060.htmlhttp://gmatclub.com/forum/pandqarei ... 09038.htmlhttp://gmatclub.com/forum/questionabou ... 08086.htmlhttp://gmatclub.com/forum/ifnisthep ... 06289.htmlhttp://gmatclub.com/forum/whatistheg ... 05746.htmlhttp://gmatclub.com/forum/ifdisapos ... 26692.htmlhttp://gmatclub.com/forum/if10252i ... 06060.htmlhttp://gmatclub.com/forum/howmanyzero ... 42479.html 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!
_________________
Reaching my Goal Gmat Score #the best is yet to come #believe



Math Expert
Joined: 02 Sep 2009
Posts: 57022

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
19 Dec 2017, 03:26
gracecud46 wrote: Bunuel wrote: FACTORIALSThis post is a part of [ GMAT MATH BOOK] created by: Bunueledited by: bb, Bunuel DefinitionThe factorial of a nonnegative 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.
 \(n!=(n1)!*n\), 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 nonnegative 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: http://gmatclub.com/forum/if60iswrit ... 01752.htmlhttp://gmatclub.com/forum/howmanyzero ... 00599.htmlhttp://gmatclub.com/forum/findthenumb ... 08249.htmlhttp://gmatclub.com/forum/findthenumb ... 08248.htmlhttp://gmatclub.com/forum/ifnisthep ... 01187.htmlhttp://gmatclub.com/forum/ifmisthep ... 08971.htmlhttp://gmatclub.com/forum/ifpisanat ... 08251.htmlhttp://gmatclub.com/forum/if10252i ... 06060.htmlhttp://gmatclub.com/forum/pandqarei ... 09038.htmlhttp://gmatclub.com/forum/questionabou ... 08086.htmlhttp://gmatclub.com/forum/ifnisthep ... 06289.htmlhttp://gmatclub.com/forum/whatistheg ... 05746.htmlhttp://gmatclub.com/forum/ifdisapos ... 26692.htmlhttp://gmatclub.com/forum/if10252i ... 06060.htmlhttp://gmatclub.com/forum/howmanyzero ... 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! 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.
_________________



Wisconsin(Madison) School Moderator
Joined: 06 Mar 2017
Posts: 192
Concentration: Operations, General Management

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
15 May 2018, 03:24
aimhigh wrote: 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 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!?



IIMA, IIMC School Moderator
Joined: 04 Sep 2016
Posts: 1365
Location: India
WE: Engineering (Other)

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
31 Jul 2018, 06:26
Bunuel chetan2u KarishmaB niks18 GMATPrepNow EMPOWERgmatRichCWhat 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?
_________________
It's the journey that brings us happiness not the destination. Feeling stressed, you are not alone!!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9535
Location: Pune, India

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
01 Aug 2018, 05:22
adkikani wrote: Bunuel chetan2u KarishmaB niks18 GMATPrepNow EMPOWERgmatRichCWhat 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? I have explained why you need 2 and/or 5 in the denominator for a terminating decimal in this post: https://www.veritasprep.com/blog/2013/1 ... thegmat/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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



IIMA, IIMC School Moderator
Joined: 04 Sep 2016
Posts: 1365
Location: India
WE: Engineering (Other)

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
01 Aug 2018, 06:59
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.
_________________
It's the journey that brings us happiness not the destination. Feeling stressed, you are not alone!!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9535
Location: Pune, India

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
02 Aug 2018, 03:31
adkikani wrote: 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. If b = 1, a/b is an integer and the question of terminating/nonterminating 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:
decimals1305.gif [ 7.14 KiB  Viewed 2405 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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



NonHuman User
Joined: 09 Sep 2013
Posts: 12007

Re: Everything about Factorials on the GMAT
[#permalink]
Show Tags
05 Aug 2019, 18:19
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.
_________________




Re: Everything about Factorials on the GMAT
[#permalink]
05 Aug 2019, 18:19



Go to page
Previous
1 2
[ 34 posts ]



