[seqfan] Re: Dividing the sum of the k leftmost digits of N by k

[Eric Angelini]

.... mmmh, I have just found 5102794 now...
Best,
É.


-----Message d'origine-----
De : Eric Angelini 
Envoyé : mercredi 26 août 2009 19:04
À : 'Sequence Fanatics Discussion list'
Objet : Dividing the sum of the k leftmost digits of N by k


Hello SeqFans, [idea coming from the recent 'average' post by Zakir]

is 978015 the biggest number N with no two same digits having the pro-
perty that when the sum of the k leftmost digits of N is divided by k
the result is always an integer?
                                               N = 978015 

- dividing the sum of the 2 leftmost digits by 2: (9+7)/2 = 8 
- dividing the sum of the 3 leftmost digits by 3: (9+7+8)/3 = 8
- dividing the sum of the 4 leftmost digits by 4: (9+7+8+0)/4 = 6
- dividing the sum of the 5 leftmost digits by 5: (9+7+8+0+1)/5 = 5
- dividing the sum of the 6 leftmost digits by 6: (9+7+8+0+1+5)/6 = 5

Many seq based on this idea could be added to the OEIS (if of interest)

The same with the rightmost digits.

(see http://www.research.att.com/~njas/sequences/A061383
     "Arithmetic mean of digits is an integer.")
Best,
É.