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Re: Everything about Factorials on the GMAT [#permalink]
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17 Apr 2014, 02:09
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Bunuel wrote: agavaqif wrote: Very nice post, indeed.is there anything about factors of factorial? Check the links in my signature below. Thankss. Actually after writing this post I found necessery info from links in your profile. Thanks again



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11 Jun 2014, 12:18
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?



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11 Jun 2014, 13:13



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Re: Everything about Factorials on the GMAT [#permalink]
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27 Aug 2014, 05:05
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I don't know the proof but I thought this would be useful:
Where a and b are integers,
(a + b)! is divisible by a!.b!



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12 Dec 2014, 18:57
Really valuable, thanks!



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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.
Hope it helps.



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05 May 2015, 08:47
If there were option give 100 kidos at once, I would give for this post. Thanks a hundred!!
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Re: Everything about Factorials on the GMAT [#permalink]
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26 May 2015, 11:51
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, 02:34
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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22 Sep 2015, 14:58
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?



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Re: Everything about Factorials on the GMAT [#permalink]
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04 Nov 2015, 03:16
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? ayushkhatriIn 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! .
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Re: Everything about Factorials on the GMAT [#permalink]
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31 May 2016, 05:35
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 Hi, When you say how many powers of 900 are in 50!, i understand you mean the powers of the prime factors of 900 in 50!. Kindly confirm? Further what do you mean by "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" Thanks



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Re: Everything about Factorials on the GMAT [#permalink]
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31 May 2016, 07:54
nishi999 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 Hi, When you say how many powers of 900 are in 50!, i understand you mean the powers of the prime factors of 900 in 50!. Kindly confirm? Yes Further what do you mean by "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" Prime factorization of 900 gives 900=\(2^{2}\)∗\(3^{2}\)∗\(5^{2}\) From this prime factorization , we can conclude that all of the prime factors 2 , 3 and 5 need to be represented twice in 900. We also know the power of 5 in 50! is 5^12 .We selected 5 and its power because among 2,3 and 5  power of 5 is least . Hence , we can say that there the 12 powers of 5 can be divided among 6 pairsThanks Hi nishi999 , Please find my responses as highlighted text in blue . 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 \(p^k\leq{n}\) Hope this help!!
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01 Aug 2016, 00:42
Please Bunuel Could you explain how you got the result below.
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



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07 May 2017, 07:31
Bunuel wrote: 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.
Can someone please explain what this sentence means?



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07 May 2017, 07:50
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In order to calculate the number of trailing zeros we need to determine the # of times 10 appears in the given factorial
However 10=5x2, in order to make a 10 both 5 and 2 need to present in equal measure. eg 5x5x2= 50 (Only 1 trailing zero as there is only one 5 and 2 to make a 10). But in 5x5x2x2=100 (Gives us 2 trailing zeros as there are 1 5s and 2s)
Now coming back to your question we need to determine the no of 5s and 2s in n!. The lower of the two (ie their power) will give the no of trailing zeros. As the power of 2 will always be higher in n! (considering its contains multiple consecutive nos) it would be sufficient to just determine the no of 5s.
In case of any doubt you could calculate the no of 2s as well and then select the lower of the two (# of 5 or 2)
Hope this clarifies !



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Re: Everything about Factorials on the GMAT [#permalink]
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19 Dec 2017, 02:15
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!



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19 Dec 2017, 02:24
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, 02:26
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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29 Jan 2018, 02:19
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[/quote]
Can someone please explain to me the above line.




Re: Everything about Factorials on the GMAT
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29 Jan 2018, 02:19



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