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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?

See The Solution Submitted by Justin    
Rating: 4.0000 (3 votes)

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Solution Proof/Explanation Comment 2 of 2 |

Stirling's approximation ln(n!) 「 n*ln(n)-n gives a good approximation of n! for large values of n (and Charlie showed they do get large)

So let's use x = 10^n*e

ln((10^n*e)!) 「 (10^n*e)ln(10^n*e) - 10^n*e
ln((10^n*e)!) 「 (10^n*e)(ln(10^n)+ln(e)) - 10^n*e
ln((10^n*e)!) 「 (n*10^n*e*ln10 + 10^n*e*1) - 10^n*e
ln((10^n*e)!) 「 n*10^n*e*ln10
ln((10^n*e)!) 「 n*10^n*e/log(e)
log((10^n*e)!) 「 n*10^n*e
Substitute back so see we got what we wanted
log(x!) 「 n*10^n*e =nx
int(log(x!)+1 「 int(n*10^n*e)+1 「 int(n*10^n*e)

The left side is len(x!) and the right is what Charlie already observed: e multiplied by 10^n rounded off.

  Posted by Jer on 2006-03-02 12:48:08
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