[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.

Franklin T. Adams-Watters

-----Original Message-----
From: Tanya Khovanova <mathoflove-seqfan at yahoo.com>

... another problem: the number formed by the first k integers is
divisible by k.

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