[seqfan] Re: Counting problem
Max Alekseyev
maxale at gmail.com
Wed Jun 17 06:08:53 CEST 2009
n Tue, Jun 16, 2009 at 11:22 PM, David Wilson<davidwwilson at comcast.net> wrote:
> How many n-digit numbers with all odd digits are divisible by 5^n?
There is exactly one such number for every n. The sequence of such
n-digit numbers is:
5, 75, 375, 9375, 59375, 359375, 3359375, 93359375, 193359375,
3193359375, 73193359375, ...
The elements of this sequence satisfy the recurrence:
a(n) = d(n)*10^(n-1) + a(n-1)
where d(n), the leading digit of a(n), is one of the odd digits 1, 3,
5, 7, or 9 (forming the complete set of residues modulo 5) and is
uniquely defined by the congruence:
d(n) == (- a(n-1) / 10^(n-1)) (mod 5).
The sequence of leading digits d(n) is:
5, 7, 3, 9, 5, 3, 3, 9, 1, 3, 7, 7, 3, 7, 7, 5, 1, 5, 7, 9, 3, 7, 3,
3, 9, 9, 9, 5, 3, 7, 1, 7, 9, 5, 7, 9, ...
Regards,
Max
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