Newsgroups: sci.math
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Date: 30 Dec 1994 22:25:51 GMT
Subject: Re: 1995 a very interesting number

Reiner.Eu...@psychol.uni-giessen.de (Reiner Euler) writes:

> Which 4-digit year "abcd" solves the conditions
>   ab|cd|abcd      (condition A)
>   abc|bcd         (condition B)  ?
> ... [except for a=b=c=d], I found only four numbers containing both
> conditions: 1248, 1664, 1995, 4998.

Reiner notes that all the numbers are very close to 10000/n, for
n=8,6,5,2.

  This is mostly because of condition B.

  If         k(abc) = (bcd),
  then      k(abcd) = 10(bcd) + k d
  then (10-k)(abcd) = 10000a + k d
  then (abcd) = 10000/((10-k)/a) + (k d)/(10-k)

Since the first term is large, the question is why should (10-k)/a be
an integer?  In most cases a=1 which suffices; certainly a <= 4 .
There are a few numbers (excluding k=0) satisfying condition B that do
not satisfy condition A: 1250, 1426, 2498, 2500, 2855, 3748, and 3750.
Only the last three have (10-k)/a not an integer.

Is there a radix in a number satisfying both conditions has (r-k)/a
not an integer?  I've only tested up to base 12, unsuccessfully.

Happy New Year.

Dan
Hoey@AIC.NRL.Navy.Mil