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June 17th, 2012, 06:54 AM   #1
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Rearrangement of 123456789

K=123456789 is a nine digits number
now we rearrange randomly the digit in K and will get 9 ! different numbers
if abcdefghi is one of this number for the sum of any adjacent three digits
(example :a+b+c ,b+c+d . c+d+e,.............g+h+i)
(i) please find the maximum value of N
Ans: max(N)=15
(ii) How many different numbers we can find satisfying the rule of (i)
Ans:124
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June 24th, 2012, 04:47 AM   #2
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Re: Rearrangement of 123456789

This is not a full solution, and I don't intend to add anything to it, and there may be better ways.

Firstly there is no maximum value for N.

There is however a minimum.

Here's some opening thoughts:

If we add all 7 equations, then we have

a+2b+3(c+d+e+f+g)+2h+i <= 7N

The minimum value of the LHS arises if we assign 9,8 to a and i, 7,6 to b and h, and the rest to c,d,e,f,g.

Doing this we end up with 88 <=7N, and hence 13<=N, since N must be an integer.

So, 13 is a lower bound for our N.

Can we construct a number where N is 13?

Well no. Why not?

If we consider the digits 9,8,7, then none of them can be in the same group of 3 adjacent digits as each other (as they would give a higher value of N). So:

If we divide our 9 digit number up into 3 lots of 3 (i.e. |abc|def|ghi|), then each of the digits 9,8,7 must be in a different cell.

We now need to allocate 1,2,3,4,5,6 so that the maximum value of those cells is as small as possible.

It turns out that this is 15, and this our new lower bound for N.

The question now is, can we construct a number where N is 15?

And the answer is, yes....
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June 25th, 2012, 12:07 AM   #3
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Re: Rearrangement of 123456789

Contrary to my previous opening sentence, here is a bit more, since ~99% of what I wrote previously is redundant.


If we consider the sum of the 1st three digits, and the sum of the 2nd three, and the sum of the third three, then each of these sums must be less than or equal to N, and their total less than or equal to 3N.

But the total is just the sum of the digits 1 to 9, and equals 45.

So 45 <= 3N and thus 15<=N.

It just remains to construct numbers where N=15, to show min(N)=15.
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June 25th, 2012, 03:27 AM   #4
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Re: Rearrangement of 123456789

Quote:
ghostwalker wrote :But the total is just the sum of the digits 1 to 9, and equals 45.

So 45 <= 3N and thus 15<=N. It just remains to construct numbers where N=15, to show min(N)=15
it should be 3N<=45 , and N <=15

for the sum of any adjacent three digits must <=15

so it shouls be max(N)=15 ,not Min(N)=15

for example :

(384 267 159) is only one situation of all rearrangements:

3+8+4=15, 8+4+2=14, 4+2+6=12, 2+6+7=15 , 6+7+1=14 , 7+1+5=13 , 1+5+9=15 (here 15 is max(N) )

How to get the second answer (124),this part is more difficult ,it must be something related to permutation

I wrote a computer program to get the answer (124)
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June 25th, 2012, 03:36 AM   #5
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Re: Rearrangement of 123456789

Quote:
Originally Posted by [color=#0000FF
Albert.Teng[/color]]so it shouls be max(N)=15 ,not Min(N)=15
N = 100 satisfies
Quote:
Originally Posted by You
if abcdefghi is one of this number for the sum of any adjacent three digits
. N = 1 doesn't.
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June 25th, 2012, 04:00 AM   #6
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Re: Rearrangement of 123456789

Quote:
Originally Posted by Albert.Teng
it should be 3N<=45 , and N <=15

for the sum of any adjacent three digits must <=15

so it shouls be max(N)=15 ,not Min(N)=15

for example :

(384 267 159) is only one situation of all rearrangements:

3+8+4=15, 8+4+2=14, 4+2+6=12, 2+6+7=15 , 6+7+1=14 , 7+1+5=13 , 1+5+9=15 (here 15 is max(N) )
At no point in your original post do you say the sum of any adjacent three digits must be <=15.

Since you're choosing any number you could choose 123456789 and here the max of 3 adjacent numbers is 24.

If all sums are less then something, then they are less than any number bigger than that something, so there is no max. to that something.

Quote:
How to get the second answer (124),this part is more difficult ,it must be something related to permutation

I wrote a computer program to get the answer (124)
Yes, I ended up with writing a program, and get 124.
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June 25th, 2012, 06:55 AM   #7
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Re: Rearrangement of 123456789

If, as you say, the digits in K can be rearranged randomly to produce any of the 9! expressible by these 9 digits, then the sum of any three consecutive digits can range from 1+2+3=6 to 7+8+9=24. So in what sense would 6 not be the minimum and 24 the maximum?

Of course, 6 and 24 won't appear in all arrangements. So you can ask mini-max and maxi-min questions, ie what it the number n such that the largest sum of any three consecutive integers is never smaller than n and what is the number m such that the smallest sum of any three consecutive integers is never larger than m. Is this what we are talking about here?
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June 25th, 2012, 06:59 AM   #8
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Re: Rearrangement of 123456789

If you let N<15 and again test it with your program ,you will find no answer,
that is why I set max(N)=15.
May be I did not make a clear definition
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June 25th, 2012, 07:34 AM   #9
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Re: Rearrangement of 123456789

Can you show an example of an arrangement where the highest sum of three consecutive digits equals 15? I'm pretty sure that the highest sum will never be lower than 16 and the lowest sum will never be higher than 14.
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June 25th, 2012, 07:55 AM   #10
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Re: Rearrangement of 123456789

(384 267 159) is only one situation of all rearrangements:

3+8+4=15, 8+4+2=14, 4+2+6=12, 2+6+7=15 , 6+7+1=14 , 7+1+5=13 , 1+5+9=15 (here 15 is max(N) )
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