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

N. J. A. Sloane njas at research.att.com
Thu Aug 27 21:50:18 CEST 2009

```Dear Seq Fans,

Yesterday the following emails arrived about a rather nice sequence.

Could someone please please submit the main sequence?  Harvey, maybe,
since it looks like you have the full list?

Any really intteresting related sequences would be welcome too!

Just because I am on "vacation" doesn't mean that I have stopped
collecting, or reading email!

Thanks

neil

Date: Wed, 26 Aug 2009 19:04:21 +0200
From: "Eric Angelini" <Eric.Angelini at kntv.be>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan]  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.

From: "Eric Angelini" <Eric.Angelini at kntv.be>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Dividing the sum of the k leftmost digits of N by k

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

Date: Wed, 26 Aug 2009 13:40:40 -0400
From: David Wilson <dwilson at gambitcomm.com>

Eric Angelini wrote:
>
>
> .... mmmh, I have just found 5102794 now...
> Best,
> É.

and you will find

5106394
5160394
7984205
8439605
8493605
8497205
9784205

and that will be it.

X-Mailer: YahooMailClassic/6.1.2 YahooMailWebService/0.7.338.2
Date: Wed, 26 Aug 2009 11:02:04 -0700 (PDT)
From: Tanya Khovanova <mathoflove-seqfan at yahoo.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

A ten digit number like that exists, and it is unique. I forgot what it is, but I can find it if I have time:
http://blog.tanyakhovanova.com/?p=31

I believe Martin Gardner wrote about it.

Date: Wed, 26 Aug 2009 11:32:32 -0700 (PDT)
From: Robert Israel <israel at math.ubc.ca>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

It can't exist, since for k=3D10, 0+1+...+9 =3D 45 is not divisible by 10.=
=20
Moreover, if my Maple program is to be believed, there are none with 8 or=
=20
9 digits.  However, there are 15 with 7 digits:
1502794
1506394
1560394
2015794
4839605
4893605
4897205
5102794
5106394
5160394
7984205
8439605
8493605
8497205
9784205

Cheers,
Robert Israel

From: "Harvey P. Dale" <hpd1 at nyu.edu>

Further to Robert Israel's post (showing that there are 15
7-digit numbers satisfying the tests), here is a complete count of the
numbers that satisfy the tests:

Number of Digits                   Total number

1                                                                9

2                                                              36

3                                                             105

4                                                             165

5                                                             233

6                                                             110

7                                                               15

That makes a grand total of 673 numbers that satisfy the tests.

Best,

Harvey P. Dale

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From: franktaw at netscape.net

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.

From: =?iso-8859-1?Q?Ignacio_Larrosa_Ca=F1estro?= <ilarrosa at mundo-r.com>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>

Tanya Khovanova wrote:
> I am very sorry.
>
> I confused it with another problem: the number formed by the first k
> integers is divisible by k.
>

Do you meant 381654729(0)?

Saludos,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa at mundo-r.com

Date: Wed, 26 Aug 2009 14:12:13 -0700 (PDT)
From: Tanya Khovanova <mathoflove-seqfan at yahoo.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

Yes, I meant that.

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