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ritula
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 Posted: Wed Mar 18, 2009 2:41 am |
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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 Source: GMAT Club Tests - hardest GMAT questions
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gmatbull
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Posted: Fri Sep 03, 2010 4:28 am |
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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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bandit
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Posted: Wed Mar 18, 2009 3:21 am |
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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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dkverma
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Posted: Fri Sep 03, 2010 3:41 am |
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C.
15! has 10*(4*5)*(8*15) = atleast 3 0's at the end
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Bunuel
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Posted: Fri Sep 03, 2010 4:44 am |
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craky
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Posted: Fri Sep 03, 2010 4:52 am |
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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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sravani
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Posted: Fri Sep 03, 2010 5:22 am |
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FQ
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Posted: Fri Sep 03, 2010 11:19 am |
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babyif19
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Posted: Sat Sep 04, 2010 7:16 am |
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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 Source: GMAT Club Tests - hardest GMAT questions
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ohfred
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Posted: Sun Sep 05, 2010 7:33 am |
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sarathy
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Posted: Sun Sep 05, 2010 8:33 am |
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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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prashantbacchewar
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Posted: Sun Sep 05, 2010 10:44 pm |
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ggallo
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Posted: Thu Sep 09, 2010 10:06 am |
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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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alexgmd
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Posted: Thu Sep 08, 2011 3:55 am |
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since 1000 = 2*2*2*5*5*5
N! should include three 5s => 5, 10 & 15
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321kumarsushant
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Posted: Thu Sep 08, 2011 4:04 am |
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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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subhajeet
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Posted: Thu Sep 08, 2011 8:03 am |
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Bowtie
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Posted: Thu Sep 08, 2011 10:09 am |
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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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viks4gmat
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Posted: Thu Sep 08, 2011 11:37 am |
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Bunuel rocks!! i used his method to get this answer in 10 sec.
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catfreak
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Posted: Thu Sep 08, 2011 11:38 am |
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rongali
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Posted: Thu Sep 08, 2011 2:56 pm |
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