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# number of terminal zeros

Author Message
Manager
Joined: 26 Dec 2008
Posts: 57
Schools: Booth (Admit R1), Sloan (Ding R1), Tuck (R1)

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05 Mar 2009, 08:27
how many zeros are there at the end of 100!

a. 10
b. 20
c. 24
d. 28
e. 30

I thought this problem was pretty interesting so sharing it here. I will reveal the answer and approach after I get a few replies.

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Current Student
Joined: 28 Dec 2004
Posts: 3292
Location: New York City
Schools: Wharton'11 HBS'12
Re: number of terminal zeros [#permalink]

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05 Mar 2009, 08:55
i get 24..

2^a 5^b ...breakdown the 100! to its prime factor, the power 5 determines the numbers of zeros..
Senior Manager
Joined: 30 Nov 2008
Posts: 479
Schools: Fuqua
Re: number of terminal zeros [#permalink]

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05 Mar 2009, 09:35
FN wrote:
i get 24..

2^a 5^b ...breakdown the 100! to its prime factor, the power 5 determines the numbers of zeros..

This sounds a better approach. But can you please elaborate how you came with the power of 5 to be 24?
Senior Manager
Joined: 30 Nov 2008
Posts: 479
Schools: Fuqua
Re: number of terminal zeros [#permalink]

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Updated on: 05 Mar 2009, 11:47
1
Never mind. I figured it out.

1. in 100!, we have 100 / 5 numbers that have 5 as the prime factor. So we get a count of 20.

2. Now 25 integer have an additoinal 5 as a factor. So 100 / 25 gives an additional 5 as the prime factor. So we get a count of 4.

Now we have the power of 5 to be 20 + 4 which is 24.

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

Simple Rule worth noting to find the number of terminating zeroes of n!

1. Divide the number n by 5^m where m ranges from 1 thru a max value x such that 5^x is less than the number.
2. Sum of all the quotients obtained is the Ans.

For in our case n = 100.

Divide 100 by 5, 5^2. Quotients are 20 and 4. So the final ans is 24.

another example for 200!.

200 / 5 = 40
200 / 25 = 8
200 / 125 = 1. (Stop at this point. 5^4 = 625 < 200 and division results zero from here onwards. )

So ans is 49.

Some good alternatives are provided in the below link.

Edit: URL was broken

Originally posted by mrsmarthi on 05 Mar 2009, 10:54.
Last edited by mrsmarthi on 05 Mar 2009, 11:47, edited 1 time in total.
Director
Joined: 01 Aug 2008
Posts: 651
Re: number of terminal zeros [#permalink]

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05 Mar 2009, 11:26
good explanation mrsmarthi . +1 for you.

Thanks.
Manager
Joined: 26 Dec 2008
Posts: 57
Schools: Booth (Admit R1), Sloan (Ding R1), Tuck (R1)
Re: number of terminal zeros [#permalink]

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05 Mar 2009, 12:19
mrsmarthi wrote:
Never mind. I figured it out.

1. in 100!, we have 100 / 5 numbers that have 5 as the prime factor. So we get a count of 20.

2. Now 25 integer have an additoinal 5 as a factor. So 100 / 25 gives an additional 5 as the prime factor. So we get a count of 4.

Now we have the power of 5 to be 20 + 4 which is 24.

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

Simple Rule worth noting to find the number of terminating zeroes of n!

1. Divide the number n by 5^m where m ranges from 1 thru a max value x such that 5^x is less than the number.
2. Sum of all the quotients obtained is the Ans.

For in our case n = 100.

Divide 100 by 5, 5^2. Quotients are 20 and 4. So the final ans is 24.

another example for 200!.

200 / 5 = 40
200 / 25 = 8
200 / 125 = 1. (Stop at this point. 5^4 = 625 < 200 and division results zero from here onwards. )

So ans is 49.

Some good alternatives are provided in the below link.

Edit: URL was broken

Right approach and answer. Good show!

Manager
Joined: 24 Feb 2007
Posts: 230
Location: nj
Re: number of terminal zeros [#permalink]

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05 Mar 2009, 13:54
1
mrsmarthi wrote:
Never mind. I figured it out.

1. in 100!, we have 100 / 5 numbers that have 5 as the prime factor. So we get a count of 20.

2. Now 25 integer have an additoinal 5 as a factor. So 100 / 25 gives an additional 5 as the prime factor. So we get a count of 4.

Now we have the power of 5 to be 20 + 4 which is 24.

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

Simple Rule worth noting to find the number of terminating zeroes of n!

1. Divide the number n by 5^m where m ranges from 1 thru a max value x such that 5^x is less than the number.
2. Sum of all the quotients obtained is the Ans.

For in our case n = 100.

Divide 100 by 5, 5^2. Quotients are 20 and 4. So the final ans is 24.

another example for 200!.

200 / 5 = 40
200 / 25 = 8
200 / 125 = 1. (Stop at this point. 5^4 = 625 < 200 and division results zero from here onwards. )

So ans is 49.

Some good alternatives are provided in the below link.

Edit: URL was broken

Just an application of the above solution

this kind of question can also be put like:

what power of 15 divides 87! exactly.

15 = 3*5

87/5 + 87/5^2 = 17 + 3 = 20

87/3 + 87/3^2 +87/3^3 +87/3^4 +.. = 29 (more than 20)

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: number of terminal zeros   [#permalink] 05 Mar 2009, 13:54
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# number of terminal zeros

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