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Discussion of Problem 1407. One-two, One-two

Please, help!!!
The biggest integer, that I find is 11111212122112, but I do not find any sequense or algo, what can I do???

Edited by author 09.01.2007 00:55
Re: Please, help!!!
Posted by KIRILL(ArcSTU) 9 Jan 2007 00:57
86 digits left for solution
Use long arithmetics
Re: Please, help!!!
Posted by it4.kp 9 Jan 2007 01:55
why 86?

my solution has 98 digits.

BTW, does anyone know what is the shortest answer?
Re: Please, help!!!
My helpprogram work very long... %(((

Edited by author 09.01.2007 08:43
Re: Please, help!!!
Posted by Burunduk1 9 Jan 2007 08:47
It's very easy to constract answer with length N... (induction)
Re: Please, help!!!
Posted by KIRILL(ArcSTU) 9 Jan 2007 11:41
Burunduk1 wrote 9 January 2007 08:47
It's very easy to constract answer with length N... (induction)

You mean that it can be solved without long numbers?
I can not find nothing better than divide number on 2^N
on every step
Reccurence!
Posted by svr 11 Jan 2007 22:57
Let H[N]- number under considaration.
Let we define integer K[N]=H[N]/(2^N);
Then K[N+1]=(m*(5^(N-1))+K[N-1])/2; where m=1 or 2;
and K[1]=1;
We solve this reccurence in long arithmetics.
In my module was bad long dividing, thank to this problem
that bug removed now.
Who can explain, how can I find the number, wich devide on 2^100???
I find 2^100 and than add 2^100 until all digits will be 1 and 2... My programm worked 2 hours, but didn't give me the answer!!!
Re: Who can explain, how can I find the number, wich devide on 2^100???
Am I right???
Re: Who can explain, how can I find the number, wich devide on 2^100???
Posted by Nikolaev Maxim 6 Feb 2007 18:06
Some number is divisible on 2^n if number which has been written down in his last n digits is divisible on 2^n.

Means we can consistently select digits in the answer.