[seqfan] Re: Counting Problem

[Tanya Khovanova]

Harvey,

You missed our previous discussion and Max's prove by induction that the answer is always 1. 

The prove is that after we take the last n-1 digits to be the previous number in the sequence, all odd possibilities for the first digit give different remainders mod 5. By pigeon hole principle, exactly one of them generates the needed number.

Tanya

--- On Wed, 6/17/09, Harvey P. Dale <hpd1 at nyu.edu> wrote:

> From: Harvey P. Dale <hpd1 at nyu.edu>
> Subject: [seqfan]  Counting Problem
> To: seqfan at list.seqfan.eu
> Date: Wednesday, June 17, 2009, 6:00 PM
> David:
> 
>  
> 
>             Is the answer
> always 1?  The numbers satisfying your
> conditions, starting with n=1, are 5, 75, 375, 9375, 59375,
> 359375, and
> 3359375.  Note that once any such number is found, the
> number for the
> next higher n has the same ending digits with a prepended
> different odd
> digit.
> 
>  
> 
>             Best,
> 
>  
> 
>             Harvey
> 
>  
> 
> -----Original Message-----
> 
> From: seqfan-bounces at list.seqfan.eu
> [mailto:seqfan-bounces at list.seqfan.eu]
> On Behalf Of David Wilson
> 
> Sent: Tuesday, June 16, 2009 11:22 PM
> 
> To: Sequence Fans
> 
> Subject: [seqfan] Counting problem
> 
>  
> 
> How many n-digit numbers with all odd digits are divisible
> by 5^n?
> 
>  
> 
> 
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