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



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Re: Everything about Factorials on the GMAT [#permalink]
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24 Sep 2013, 00:35



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




Re: Everything about Factorials on the GMAT
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