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Re: least value of N (m09q33) [#permalink]
03 Sep 2010, 04:28

7

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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]
05 Sep 2010, 08:33

1

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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]
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.

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