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        riddles >> medium >> Numerically Attractive 
        
(Message started by: ThudanBlunder on May 26^th, 2010, 6:07pm)

Title: Numerically Attractive
Post by ThudanBlunder on May 26^th, 2010, 6:07pm

Consider a 4-digit number whose digits are not all equal. With these four digits form the larget possible number M and the smallest possible number m. Calculate M - m and replace the original 4-digit number with the result. Prove that repeating this procedure always arrives at the same number.

Title: Re: Numerically Attractive
Post by Noke Lieu on May 26^th, 2010, 8:08pm

I'm assuming that there's a missing eventually int here.

Given a > b > c > d
M=(1000a+100b+10c+d)
m= (1000d + 100c +10b +a)

M-m= 9(111a + 10b - 10c - 111d) =
9(100(a-d)+10(a+b-d-c)+(a+d)) = n

9(100(a+d)+10(a+b-d-c)+(a-d)) = N

N-n = 9(200d + d)

hang on, that doesn't work... :-X

Title: Re: Numerically Attractive
Post by towr on May 27^th, 2010, 2:26am

on 05/26/10 at 20:08:00, Noke Lieu wrote:

I'm assuming that there's a missing eventually int here.

To give you an idea how long it takes, here's a table that shows how many numbers take a given number of steps:

#steps #nums
0 1
1 356
2 519
3 2124
4 1124
5 1379
6 1508
7 1980
>7 0

[edit]If you don't allow numbers to start with 0, you have to exclude multiples of 1000 as well as multiples of 1111 (you get 0 in both cases, rather than the intended 'attractor'). And the table becomes:

#steps #nums
0 1
1 287
2 600
3 813
4 380
5 396
6 1296
7 1365
8 1341
9 785
10 1235
11 483
>11 0
[/edit]

[edit2]Oh, and of course this assumes you add extra zeros when M-m < 1000. Otherwise it becomes more complicated still.[/edit2]

Title: Re: Numerically Attractive
Post by ThudanBlunder on Aug 1^st, 2010, 7:06am

[hide]Let
M = abcd
and
m = dcba
where
a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif b
c http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif d
a > d

There are two cases to consider:

1) b > c
Subtracting,
a b c d
- d c b a
---------------------------------------------
[a - d] [b - 1 - c] [10 + c - 1 - b] [10 + d - a]
---------------------------------------------

The 1st and 4th digits of the result sum to 10
and
The 2nd and 3rd digits of the result sum to 8

2) b = c
Subtracting,
a b c d
- d c b a
---------------------------------------------
[a - 1 - d] [10 + b - 1 - c] [10 + c - 1 - b] [10 + d - a]
---------------------------------------------

In this case,
the 1st and 4th digits of the result sum to 9
and
the 2nd and 3rd digits of the result sum to 18

Case 1 gives rise to 25 numbers of the required form
and
Case 2 to 5 such numbers, where order of digits is not important for forming the next M and m.

These all quickly converge on the numerical attractor, 6174.[/hide]