# [seqfan] Dividing the k leftmost digits of N by k

franktaw at netscape.net franktaw at netscape.net
Wed Aug 26 22:46:14 CEST 2009

```This problem interests me, as generalized to arbitrary base b.

First, obviously, the last digit must be 0 for any b.

There are no solutions for b odd: for k = b-1, the number must be a
permutation of the digits 1 through b-1.  But the sum of the digits in
base b is congruent to the the number modulo b-1; and when b-1 is even,
this is congruent to (b-1)/2.

For any even b, the b/2 digit must be b/2, since 0 is reserved for the
final digit.  More generally, for any divisor d of b, the digit
positions divisible by d must have digits that are a multiple of d; and
this exhausts the multiples of d, so that any non-multiple of d must
have a digit value that is not a multiple of d.

See A111456, which enumerates the known values (1 in base 2, 2 in base
4, 3 in base 6, 1 in base 10 -- 3816547290, and 1 in base 14).  It
seems likely that there are no more values.