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Topic: Numerically Attractive (Read 1346 times) |
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ThudnBlunder
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Numerically Attractive
« on: May 26th, 2010, 6:07pm » |
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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.
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Noke Lieu
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Re: Numerically Attractive
« Reply #1 on: May 26th, 2010, 8:08pm » |
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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... 
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| « Last Edit: May 26th, 2010, 8:08pm by Noke Lieu » |
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towr
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Re: Numerically Attractive
« Reply #2 on: May 27th, 2010, 2:26am » |
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on May 26th, 2010, 8:08pm, Noke Lieu wrote:| I'm assuming that there's a missing eventually int here. |
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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]
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| « Last Edit: May 27th, 2010, 3:17am by towr » |
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ThudnBlunder
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Re: Numerically Attractive
« Reply #3 on: Aug 1st, 2010, 7:06am » |
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Let
M = abcd
and
m = dcba
where
a  b
c 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.
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