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# How many zeros does 1000! end with? 100 148 250 248 200

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How many zeros does 1000! end with? 100 148 250 248 200 [#permalink]  19 Mar 2008, 03:16
How many zeros does 1000! end with?

100

148

250

248

200
CEO
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Re: factorial and exp [#permalink]  19 Mar 2008, 03:31
Expert's post
1. each of zeros are defined as 2*5.
2. It is obviously that 1000! contains fewer the number of 5 than the number of 2. Therefore, the number of 5 determined the number of zeros.

(1000-5)/5+1=200
(1000-25)/25+1=40
(1000-125)/125+1=8
(625-625)/625+1=1

N=200+40+8+1=249
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Re: factorial and exp [#permalink]  19 Mar 2008, 03:33
walker wrote:
1. each of zeros are defined as 2*5.
2. It is obviously that 1000! contains greater the number of 5 than the number of 2. Therefore, the number of 5 determined the number of zeros.

(1000-5)/5+1=200
(1000-25)/25+1=40
(1000-125)/125+1=8
(625-625)/625+1=1

N=200+40+8+1=249

how did you know that it contains more 5 than 2?
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Kudos [?]: 1496 [0], given: 353

Re: factorial and exp [#permalink]  19 Mar 2008, 03:37
Expert's post
kazakhb wrote:
how did you know that it contains more 5 than 2?

Sorry, it have to be "It is obviously that 1000! contains fewer the number of 5 than the number of 2"
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Manager
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Re: factorial and exp [#permalink]  19 Mar 2008, 03:41
walker wrote:
1. each of zeros are defined as 2*5.
2. It is obviously that 1000! contains fewer the number of 5 than the number of 2. Therefore, the number of 5 determined the number of zeros.

(1000-5)/5+1=200
(1000-25)/25+1=40
(1000-125)/125+1=8
(625-625)/625+1=1

N=200+40+8+1=249

sorry for dumb question, could you explain logics behind the calculation you have made
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Kudos [?]: 1496 [0], given: 353

Re: factorial and exp [#permalink]  19 Mar 2008, 03:52
Expert's post
kazakhb wrote:
sorry for dumb question, could you explain logics behind the calculation you have made

Nb----X----X----X----X----X----X----X----Ne

Nb - is the first number in sequence
Ne - is the last number in sequence
d - is the step

The total number = (Ne-Nb)/d+1

I use this formula to avoid silly mistakes.

BTW, my answer is not correct
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Director
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Re: factorial and exp [#permalink]  19 Mar 2008, 04:18
I say 148 ...but its just a strategic guess.. I thought 100 is way to low and 250 to high.

I basically rephrased to say..how many factors of 10 are in 1000!?
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Re: factorial and exp [#permalink]  19 Mar 2008, 06:57
walker wrote:
1. each of zeros are defined as 2*5.
2. It is obviously that 1000! contains fewer the number of 5 than the number of 2. Therefore, the number of 5 determined the number of zeros.

(1000-5)/5+1=200
(1000-25)/25+1=40
(1000-125)/125+1=8
(625-625)/625+1=1

N=200+40+8+1=249

Another method I use here is to just divide 1000! by the powers of 5.

1000!/5 = 200

1000!/25 = 40

1000!/125 =8

1000!/625 ~1

All adds up to 249...??? Perhaps I we round up on the last one?
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Re: factorial and exp [#permalink]  19 Mar 2008, 08:41
i am getting 249 as well..

ans choices are off..
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Re: factorial and exp [#permalink]  19 Mar 2008, 10:00
walker wrote:
1. each of zeros are defined as 2*5.
2. It is obviously that 1000! contains fewer the number of 5 than the number of 2. Therefore, the number of 5 determined the number of zeros.

(1000-5)/5+1=200
(1000-25)/25+1=40
(1000-125)/125+1=8
(625-625)/625+1=1

N=200+40+8+1=249

why did you do [(1000 - 25)/5 + 1 ]

i do:

1000/5 = 200
1000/25 = 40
1000/125 = 8
1000/625 = 1
so total = 249
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Re: factorial and exp [#permalink]  19 Mar 2008, 10:32
Expert's post
GMAT TIGER wrote:
why did you do [(1000 - 5)/5 + 1 ]

i do:

1000/5 = 200
1000/25 = 40
1000/125 = 8
1000/625 = 1
so total = 249

to avoid some GMAT traps

For example, n is the product of 4*5*...999 and m is the product of 5*6*...1000
In both cases we have range = 995=(999-4)=(1000-5).
n contains 248 zeros but m contains 249 zeros
the formula "The total number = (Ne-Nb)/d+1" helps to avoid this trap.
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Re: factorial and exp [#permalink]  10 Jan 2012, 13:57
walker wrote:
1. each of zeros are defined as 2*5.
2. It is obviously that 1000! contains fewer the number of 5 than the number of 2. Therefore, the number of 5 determined the number of zeros.

(1000-5)/5+1=200
(1000-25)/25+1=40
(1000-125)/125+1=8
(625-625)/625+1=1

N=200+40+8+1=249

Hm, I dont see how you got 249 using this formula.
(1000-5)/5+1 = 165 ?!

Could you please explain how that formula avoids traps?

Thanks a lot!
Re: factorial and exp   [#permalink] 10 Jan 2012, 13:57
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