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Title: Single Digit Series Post by Wonderer on Dec 23rd, 2005, 3:21am There’s a series of single digit numbers. The first 6 numbers are 1,0,1,0,1,0. After these, every number in the series equals to the right most digit of the sum of the previous 6 numbers. i.e. 1,0,1,0,1,0,3,5,0,9,8,5,0,…… Question: Is it possible to ever see the following sequence in this series? 0,1,0,1,0,1 Justify your answer! Well, you are not supposed to write a program, but I can’t really stop you from doing that. However, you need to prove your answer mathematically; your program result doesn’t count. :P |
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Title: Re: Single Digit Series Post by Barukh on Dec 23rd, 2005, 3:44am on 12/23/05 at 03:21:31, Wonderer wrote:
Well, that's actually answers your question, doesn't it? ;D |
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Title: Re: Single Digit Series Post by Grimbal on Dec 23rd, 2005, 3:58am What if the program outputs a mathematical proof? Not a difficult proof, just a proof with many many steps. For instance it could make explicit all the terms up to the point where it reaches 0,1,0,1,0,1 or 1,0,1,0,1,0. |
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Title: Re: Single Digit Series Post by Wonderer on Dec 23rd, 2005, 4:11am on 12/23/05 at 03:58:35, Grimbal wrote:
That would be ok. hehe.. |
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Title: Re: Single Digit Series Post by Barukh on Dec 23rd, 2005, 4:12am on 12/23/05 at 03:58:35, Grimbal wrote:
I somehow have a feeling that such a program will never stop. |
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Title: Re: Single Digit Series Post by Wonderer on Dec 23rd, 2005, 4:23am on 12/23/05 at 04:12:26, Barukh wrote:
Me too ;D So, try not to use a program. Do it with a pen, a sheet of paper and your wisdom. |
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Title: Re: Single Digit Series Post by towr on Dec 23rd, 2005, 4:31am There's a finite number of states, at most 1000000. And because it's deterministic each state can only lead to one other state. So once you have a cycle, you have proved no states except the ones leading to it, and those in the cycle itself, are possible. |
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Title: Re: Single Digit Series Post by Grimbal on Dec 23rd, 2005, 7:38am on 12/23/05 at 04:12:26, Barukh wrote:
I don't see why it shouldn't. Given six terms, there is a unique digit that can come before. That means two sequences can not join and become identical, and therefore, the sequence can enter a loop only by restarting the sequence from the start, with 1,0,1,0,1,0. Since the number of states is finite, it must loop, so it must eventually get back to 1,0,1,0,1,0. The only question is whether it will pass thru 0,1,0,1,0,1 before ending the loop or not. But it will definitely stop and give an answer. |
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Title: Re: Single Digit Series Post by Aryabhatta on Dec 24th, 2005, 12:55am It can never become 0,1,0,1,0,1 Here is why... [hideb] Instead of modulo 10, consider modulo 5. (if acheiving 0,1,0,1,0,1 is possible for 10, it _must_ be possible for 5) We show that it is not possible for 5. We are working in the field F5 Consider the 6x6 matrix A = [0 1 0 0 0 0] [0 0 1 0 0 0] [0 0 0 1 0 0] [0 0 0 0 1 0] [0 0 0 0 0 1] [1 1 1 1 1 1] Determinant of A is -1. let the sequence which we have be x1, x2, .... Let Mn be the 6x6 matrix [xn+1 xn+2 xn+3 xn+4 xn+5 xn+6] [xn+2 xn+3 xn+4 xn+5 xn+6 xn+7] [xn+3 xn+4 xn+5 xn+6 xn+7 xn+8] [xn+4 xn+5 xn+6 xn+7 xn+8 xn+9] [xn+5 xn+6 xn+7 xn+8 xn+9 xn+10] [xn+6 xn+7 xn+8 xn+9 xn+10 xn+11] Notice that A*Mn= Mn+1 So Determinant(Mn+1) = - Determinant(Mn) We have M0 = [1, 0, 1, 0, 1, 0] [0, 1, 0, 1, 0, 3] [1, 0, 1, 0, 3, 0] [0, 1, 0, 3, 0, 0] [1, 0, 3, 0, 0, 4] [0, 3, 0, 0, 4, 3] Now Det(M0) = 84 = 4 (working in F5) Thus for any i, det (Mi) = 4 or 1. Suppose 0,1,0,1,0,1 were possible to achieve.. then for some k Mk = [0, 1, 0, 1, 0, 1] [1, 0, 1, 0, 1, 3] [0, 1, 0, 1, 3, 1] [1, 0, 1, 3, 1, 1] [0, 1, 3, 1, 1, 2] [1, 3, 1, 1, 2, 3] But the determinant value of this is -117 = 3 (mod 5) which is not 4 or 1. Hence, 0,1,0,1,0,1 isn't possible to achieve! [/hideb] |
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Title: Re: Single Digit Series Post by Barukh on Dec 24th, 2005, 9:58am Nice solution, Aryabhatta! :D |
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Title: Re: Single Digit Series Post by Aryabhatta on Dec 24th, 2005, 10:17am Thanks Barukh. :D I wonder what Wonderer had in mind... |
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