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Factorial sequence (Posted on 2006-03-02) Difficulty: 3 of 5
Factorials exhibit an interesting trait. The minimum value needed for the length of x! to reach nx, where n is a positive integer, forms an interesting sequence. Let len(x!) = int(log(x!))+1 to account for the extra digit.

What is the relationship between len(x!) and x?

What is the smallest number such that the len(x!)>=8x?

How can I approximate when len(x!) first exceeds/equals nx?

  Submitted by Justin    
Rating: 4.0000 (3 votes)
Solution: (Hide)
Part 1: len(x!) is approximately x*len(x/e)

Part 2: 271828170

Part 3: As n gets bigger, we gain more accuracy in our approximation. Our approximation of when len(x!)>=nx is xlog(x/e), where log(x/e) is approximately n. So x is approximately 10^n*e where nx is the number of digits in x!, or, n is simply the multiplier.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionProof/ExplanationJer2006-03-02 12:48:08
Solutioncomputer explorationCharlie2006-03-02 11:19:50
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