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This is simple factorial equation question.

How do you find the largest n satisfying n! < 1000?

(Edit)

Actually, I want to find some other logic other than brute force.

For example,

How about the largest n for $n!<10^{64}$?

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Well, did you try computing $5!,6!,7!$? –  Pedro Tamaroff Oct 5 '13 at 23:26
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I want to find a logical steps to find the n, other than brute force way. –  Hanasima Oct 5 '13 at 23:30
    
@Hanasima, you are on the right path. only if there was an inverse factorial or inverse $\Gamma$ function this question would have been easy. –  Arjang Oct 5 '13 at 23:31
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@Hanasima Well, $n!>2^n$ if $n>4$, and $2^{10}=1024$, so you can look at $n=1,2,3,4,5,6,7,8,9$. Is that good enough? It isn't really brute forcing. One the other hand $6!=720$ so clearly...? Have you never computed a Taylor series up to the $6$th term? :) –  Pedro Tamaroff Oct 5 '13 at 23:32
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@Hanasima: The brute force way is plenty logical. No sense trying to make an easy problem hard! –  Hurkyl Oct 5 '13 at 23:34
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2 Answers

If you are looking for non brute force methods of dealing with factorials, then probably what you want is Stirling's approximation: http://en.wikipedia.org/wiki/Stirling%27s_approximation

However, this approximation is really only helpful for large $n$; in your case, I'd say brute force is the best option.

Personally, I recall that $5! = 120$, after which $6! = 720$, and you know $7! > 1000$.


In response to the edit: To consider when $n!$ exceeds $10^{64}$ is to consider when $n!$ has at least sixty four digits; this sort of question can be broached using Stirling's Approximation combined with a base-$10$ log. Probably you could work this out yourself, but for more details, see here.

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Thanks. Now, I'm looking how to use the approximation to solve the original example. –  Hanasima Oct 5 '13 at 23:50
    
@Hanasima: Okay, I have added additional details for your subsequent edit. –  Benjamin Dickman Oct 6 '13 at 2:52
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There is a table of the first factorials at http://www.tsm-resources.com/alists/fact.html that can help you.

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that is the same as brute force, isn't there an inverse factorial function somewhere? –  Arjang Oct 5 '13 at 23:33
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@Arjang That table is the "inverse factorial function" (for natural numbers at least). –  Pedro Tamaroff Oct 5 '13 at 23:35
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You can get good approximations by inverting Stirling's theorem, but there will always be problems near a large factorial $\pm 1$. –  marty cohen Oct 5 '13 at 23:36
    
Thanks for the link. It would be helpful to find n for upto a certain number. –  Hanasima Oct 5 '13 at 23:50
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