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least value of N (m09q33)

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least value of N (m09q33) [#permalink]

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What is the least \(N\) such that \(N!\) is divisible by 1000?

(A) 8
(B) 10
(C) 15
(D) 20
(E) 25

[Reveal] Spoiler: OA
C

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New post 18 Mar 2009, 03:21
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Ans C
1000= 5^3 * 2^3

Least Number that has 5^3 * 2^3 is 15!
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New post 03 Sep 2010, 03:41
C.
15! has 10*(4*5)*(8*15) = atleast 3 0's at the end
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Re: least value of N (m09q33) [#permalink]

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New post 03 Sep 2010, 04:28
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1000 = 2^3 x 5^3
so, we are looking for a number N such that N! contains at least 2^3 x 5^3 as factors.
A. 8! = 2^3.7.6.5.4! .... only 2^3.5 wrong
B. 10! = (2*5).9.8.7.6.5.4!...only 2^3.5^2 wrong, we need an additional 5.
C. 15! = (3*5).(2*7).13.(2*6).11.(2*5).9.8! 2^3.5^3 is available in 15!...corect
D. 20! we looking for the least- 20! is not necessary
E. 25! same as option D
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Re: least value of N (m09q33) [#permalink]

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New post 03 Sep 2010, 04:44
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ritula wrote:
What is the least \(N\) such that \(N!\) is divisible by 1000?

(A) 8
(B) 10
(C) 15
(D) 20
(E) 25

[Reveal] Spoiler: OA
C

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\(n!\) to be divisible by 1,000 it must have 3 trailing zeros. So \(n\) must have 3 factors of 5 in it, so the answer is 15 as 15/5=3.

Answer: C.

For more on trailing zeros see: everything-about-factorials-on-the-gmat-85592.html

Hope it helps.
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Re: least value of N (m09q33) [#permalink]

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New post 03 Sep 2010, 04:52
Since 1000 = 2^3 x 5^3 we need 3 times 5 to do that job. 5*10*15 makes this. simple. C
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Re: least value of N (m09q33) [#permalink]

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New post 04 Sep 2010, 07:16
1000= 2^3 * 5^3

So you need at least a number that will repeat the 5 three times for you.....C


ritula wrote:
What is the least \(N\) such that \(N!\) is divisible by 1000?

(A) 8
(B) 10
(C) 15
(D) 20
(E) 25

[Reveal] Spoiler: OA
C

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Re: least value of N (m09q33) [#permalink]

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New post 05 Sep 2010, 08:33
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For such questions, always work through the answers choice from lowest to highest i.e. A to E. The moment you get an answers that matches, stop and mark that as the answer and move on.

1000 = 2^3 x 5^3
N! has to contain 2^3 x 5^3 as factors.
A) 8! --> only 2^3 * 5^1 --> wrong
B) 10! --> only 2^3 * 5^2 --> wrong
C) 15! = --> 2^3 * 5^3 --> BINGO
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Re: least value of N (m09q33) [#permalink]

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New post 09 Sep 2010, 10:06
For the N! to be divisible by 1,000, it needs 3 5's and 3 2's in it (5^3 and 2^3). We can just look at 5's since 2's will be more abundant. There are some good posts on divisibility by factors of 10 that work out this more clearly.

As we go up from 1!, we would get a 5 in each of: 5, 10, and 15. If you want to check there are three 2's by the time you get to 4! so we're covered. 15! will be the first that is divisible by 1,000.

It might be tempting to look at 10! since you get above 100 so quickly, but 10! if 3,628,800, which /1,000 is not an integer (done in excel, not needed on the test or if you understand the concepts.
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Re: least value of N (m09q33) [#permalink]

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New post 08 Sep 2011, 03:55
since 1000 = 2*2*2*5*5*5
N! should include three 5s => 5, 10 & 15
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Re: least value of N (m09q33) [#permalink]

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New post 08 Sep 2011, 04:04
easy..
ans is C.
1000=5*5*5*2*2*2
2 wont create any problem , we need to look for 5.
min no in which we will get three 5 is 15 (5,10,15).

15!=1*2*3*4*5*6*7*8*9*10*11*12*13*14*15
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Re: least value of N (m09q33) [#permalink]

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New post 08 Sep 2011, 10:09
C.1000= 5^3 * 2^3 so you need 3 5's in your answer, which C gives you. or that's at least my thought process
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Re: least value of N (m09q33) [#permalink]

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New post 08 Sep 2011, 11:37
Bunuel rocks!! i used his method to get this answer in 10 sec.
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Re: least value of N (m09q33) [#permalink]

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New post 08 Sep 2011, 14:56
easy one C is the answer
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Re: least value of N (m09q33) [#permalink]

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New post 25 Oct 2011, 21:00
sarathy wrote:
For such questions, always work through the answers choice from lowest to highest i.e. A to E. The moment you get an answers that matches, stop and mark that as the answer and move on.

I would add that for "least" questions this should be done. The opposite should be done for "greatest".
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Re: least value of N (m09q33) [#permalink]

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New post 10 Sep 2012, 08:09
What is the least \(N\) such that \(N!\) is divisible by 1000?

we need "three zeros" in the factorial multiplication as 1000 has it in.

(A) 8
it will have only one zero - 2 & 5 will multiply and get one zero- incorrect

(B) 10
it will give two zeros- 2&5 will give one zero ; 10 will have one - incorrect

(C) 15
it will give 3 zeros - 2&5;10;12&15 -correct

(D) 20
more than 3 zeros

(E) 25
more than 3 zeros
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Re: least value of N (m09q33) [#permalink]

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New post 10 Sep 2012, 11:06
Bunuel that is awesome. Great time saver. Thank you kindly.
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Re: least value of N (m09q33) [#permalink]

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New post 10 Nov 2016, 11:16
Here for n! to be divisible by 1000 it must have at-least 3 fives as the number of 2's will always be ample.
To have 3 fives n must be 15.
Hence C
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Re: least value of N (m09q33)   [#permalink] 10 Nov 2016, 11:16
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