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back to boardPlease, 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!!! 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!!! It's very easy to constract answer with length N... (induction) Re: Please, help!!! 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??? 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. |
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