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Intern
Joined: 22 Jul 2019
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why trailing zeros in factorial
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27 Jul 2019, 04:01
Just wondering what is the use of finding trailing zeros in a factorial? i tried the internet and every where they just explain how to do it.. but i want to know the practical use of it.
thanks,



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why trailing zeros in factorial
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27 Jul 2019, 04:45
Clarify your question. What do you mean by "trailing zero"? If you are referring to "zero factorial", its value by definition is one. Multiplication by one will not change the value of original quantity.
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Re: why trailing zeros in factorial
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27 Jul 2019, 04:58
Hovkial wrote: Clarify your question. What do you mean by "trailing zero"? If you are referring to "zero factorial", its value by definition is one. Multiplication by one will not change the value of original quantity.
Posted from my mobile device Hovkial, trailing zeros in a n! is defined as\(n/5+n/5^2+n/5^3\)...... \(n/5^k\)where \(k <=n\).. I was wondering the use of this in practical applications. I always understand concepts if i can attribute it to some practical application. thanks



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Re: why trailing zeros in factorial
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29 Jul 2019, 09:49
Not particularly useful, don't worry about it.
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why trailing zeros in factorial
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29 Jul 2019, 09:59
yessvee wrote: Hovkial wrote: Clarify your question. What do you mean by "trailing zero"? If you are referring to "zero factorial", its value by definition is one. Multiplication by one will not change the value of original quantity.
Posted from my mobile device Hovkial, trailing zeros in a n! is defined as\(n/5+n/5^2+n/5^3\)...... \(n/5^k\)where \(k <=n\).. I was wondering the use of this in practical applications. I always understand concepts if i can attribute it to some practical application. thanks I still do not understand what you mean by the term "trailing zeroes in n factorial". Your example is that of a series or a sequence. Series and sequences are very important to model events, e.g., in science, business, mathematics, etc and many applied disciplines. Other than that, I have no clue about your question. Cheers.



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Re: why trailing zeros in factorial
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29 Jul 2019, 10:07
Hovkial wrote: yessvee wrote: Hovkial wrote: Clarify your question. What do you mean by "trailing zero"? If you are referring to "zero factorial", its value by definition is one. Multiplication by one will not change the value of original quantity.
Posted from my mobile device Hovkial, trailing zeros in a n! is defined as\(n/5+n/5^2+n/5^3\)...... \(n/5^k\)where \(k <=n\).. I was wondering the use of this in practical applications. I always understand concepts if i can attribute it to some practical application. thanks I still do not understand what you mean by the term "trailing zeroes in n factorial". Your example is that of a series or a sequence. Series and sequences are very important to model events, e.g., in science, business, mathematics, etc and many applied disciplines. Other than that, I have no clue about your question. Cheers. if you look at the gmatclub math book, there is a section that shows a way to count the zeros at the end of a number. i was wondering where this will be used in real life. .. Trevor mentioned it is not used in real life except as an exercise math problems.



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why trailing zeros in factorial
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Updated on: 11 Aug 2019, 02:09
If I understood your question correctly I think you are trying to find the number of 0's at the end of the result of the factorial, hence how many 10's you can take out of the result. Think of it as counting the number of 10's you can divide out of the factorial, 10 = 5*2 so we need a pair of 5 and a 2 for each ten. Since factorials will always have more 2's than 5's, the question can be furthered simplified to how many 5's are there in the prime factorization of the factorial? Now typically you will have one every five numbers with the exception of numbers like 25 and 125, which are 5*5 and 5*5*5 respectively. For 100! there are 20 multiples of 5 from 5, 10, 15 ... 95, 100. Then we need to identify the multiples which contain more than one 5. The numbers are: 25 (5*5), 50 (5*5*2), 75 (5*5*3), and 100 (5*5*2*2). There are 4 numbers with an additional five in its prime factorization, so in total we have 20 + 4 = 24 copies of fives. This means that there are 24 trailing zero's for 100! Hopefully, this answers what you were asking for!
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Re: why trailing zeros in factorial
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10 Aug 2019, 19:11
TestPrepUnlimited wrote: If I understood your question correctly I think you are trying to find the number of 0's at the end of the result of the factorial, hence how many copies of 10 you can take out of the result.
Think of it as counting the number of 10's you can divide out of the factorial, 10 = 5*2 so we need one copy of five and one copy of two for each ten. Then the question is just how many copies of fives there are in the number, since there are plenty of twos to take out. Now typically this means every 5 numbers there will be a factor of 10 however don't forget 25 and 125 for example have more than 1 copy of five. To tackle that, we can count the factors in layers.
For 100! there are 20 copies of fives from 5, 10, 15 ... 95, 100. This is our first layer. The second layer is 25, 50, 75, 100. There are 4 numbers with an additional copy of five, so in total we have 20 + 4 = 24 copies of fives. Finally this results in 24 trailing zero's for 100!. If the question is about 500! instead we would need to add a third layer for each multiple of 125. The formula stated above is not useful for application however the logic and steps taken to reach to that conclusion is within GMAT difficulty. Hopefully that answers what you were asking for! thanks for your explanation.. much appreciated. is this used only in testing. is there an application where i can use the number of zeros. I am not sure if i am making sense.. rgds, sv




Re: why trailing zeros in factorial
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