Firsts Digits of n!

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

Show that there exists a natural number $n$ so that the first digits of $n!$ is $2001$

**Posted:** Tue Jan 11, 2005 10:19 pm

ondrob

I'd very like to see a solution to this problem. I've seen similar problem but instead of 2001 there was another number and I couldn't solve it... Sad

**Posted:** Thu Jan 13, 2005 5:46 am

_________________
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pbornsztein

The following is due to my friend Bruno Langlois, which will probably join us soon Smile

It solves the given problem and more general ones.

Let $\mathbb {T} = \mathbb {R} / \mathbb {Z}$ be the torus.
If $(s_n)$ is a real sequence, let $( \Delta s)_n = s_{n+1}-s_n$
Let $\Delta ^2 = \Delta o \Delta.$

Each of the four following claims is not difficult to prove (but quite boring to write) :

Claim 1 :
If $s$ is a real sequence which has limit $+\infty$ such that $\Delta s$ has limit $0$ , then $s$ is dense in $\mathbb {T}$ .

Claim 2:
If $s$ is a real sequence such that $\Delta ^2 s$ has limit $0$ and $\Delta s$ is dense in $\mathbb {T}$ , Then $s$ is dense is $\mathbb {R}$ .

Claim 3 :
If $w$ is a real sequence such that $\Delta w$ has limit $+\infty$ and $\Delta ^2 w$ has limit $0$ , then $w$ is dense in $\mathbb {T}$ .
(Just use claim 1 for $s = \Delta w$ to see that $\Delta w$ is dense in $\mathbb {T}$ and use claim 2)

Claim 4 :
If the sequence $(\log_{10} (u_n))$ is dense in $\mathbb {T}$ then for all positive integer $k$ , there exists $n$ such that the decimal representation of $k$ begins with the decimal representation of $k$ .

Thus, you just have to use it for $u_n = n!$ , starting with claim 3 for $w_n = \log_{10} (u_n)$ and then claim 4.

Pierre.

**Posted:** Thu Jan 13, 2005 6:52 am

Gyan

Similar argument but a little more intutive -

we will prove that there exist a N such that ( 10^(2N)) ! starts with 2001.. that is n = 10^2N

All logs based 10:

(1/2) log (2pi) + (n+1/2) (2N) - n log (e) + (1/(12n+1)) log e <log(n!) < (1/2) log (2pi) +(n+1/2)(2N) - nlog e + (1/12n)

We are only interested in the fractional part (mantissa), and have to prove that there exist a N where it is between log(2.001) and log (2.002)

Since (log e ) is irrational mantissa of (n log e) can fall in any small range we desire..

pbornsztein

**Posted:** Thu Jan 13, 2005 2:42 pm

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

**Posted:** Thu Jan 13, 2005 9:03 pm

Gyan

Yes, there was a typo, I should have said log(n!) (Corrected in the original message)
but anyway that was the famous Stirlings Approximation
(I just did a google and found it here:

http://mathworld.wolfram.com/StirlingsApproximation.html

One of the easy way to prove that, is to write the n! = int ( 0 to infinity) (x^n * e ^(-x) dx
If you take the function (x^n * e^(-x) ) as (exp ( f(x))) and expand f(x) near its maximum (which happens to be when x=n) so you get something like

f(x-n) + (1/2) f'' (x-n)^2) + ... Ignore the other terms and take the integral. etc..

**Posted:** Fri Jan 14, 2005 9:43 am

Gyan

This problem has come to life again here

It is in Singapore Math Olympiad (for 9th-10th graders) .. and I just realized a nice proof which does not require higher-level math. If there is an interest, I may post the solution in the other thread.

**Posted:** Wed Jul 20, 2005 11:44 am

Bojan Basic

**Posted:** Thu Aug 04, 2005 5:26 am

_________________
- Why do mathematicians get Christmas and Halloween mixed up?
- Because Dec. 25 = Oct. 31.

mathwizarddude

**Posted:** Sat Jul 12, 2008 8:03 am