# [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.

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

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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