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What digit would you put in place of question mark in the below sequence and why?

0110100011111?0011000011011

P.S. Hint: 0=1, 1=0,...

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closed as too broad by Aza Aug 11 '16 at 16:14

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Not prime: $55035419=2896601^{1}19^{1}$; $55043611=419^{1}383^{1}7^{3}$ $\endgroup$ – Jonathan Allan Aug 11 '16 at 13:24
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    $\begingroup$ The missing digit should be 0 or 1. lol $\endgroup$ – Arulkumar Aug 11 '16 at 13:41
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    $\begingroup$ @Arulkumar: We don't know that. There are 27 digits, which might suggest a grouping of 9 letters, each represented by a three-digit ternary number. $\endgroup$ – M Oehm Aug 11 '16 at 13:43
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    $\begingroup$ I would place a 0 - because it looks good. Don't you think you need to narrow down the puzzle a bit? As it is worded now, there is not a single clue of what would be "correct". Say "...because I wanted the number to be like Pi" or "...because I wanted the number NOT to be Pi" are equally valid! $\endgroup$ – BmyGuest Aug 11 '16 at 15:39
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    $\begingroup$ Hello! I've put this question on hold for now as too broad, because I'm not sure this question sufficiently limits the scope of possible answers. If you can find a way to rebuild this puzzle so that it has a limited number of objectively correct answers, please feel free to edit it! $\endgroup$ – Aza Aug 11 '16 at 16:16
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Might be number is

0

Because

bitwise XOR operation

011 ^ 010 = 001

111 ^ 100 = 011

000 ^ 011 = 011

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  • $\begingroup$ No, it was not intended! Moreover, in your logic I see an imbalance in the number of digits on both the sides, i.e., (3+2+1+7)=(4+3+0+3+3), which I would avoid in any case! $\endgroup$ – Hitesh Dholaria Aug 11 '16 at 16:03
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    $\begingroup$ @Hitesh I have modified my question. $\endgroup$ – user26522 Aug 12 '16 at 10:01
  • $\begingroup$ @Anjan seems correct! +1 $\endgroup$ – z100 Aug 12 '16 at 11:36
0
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I guess this is too simple thought..

But I go with

0

because

the parity bit is 1 and there are already 13 '1'

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  • $\begingroup$ I rather posted this as comment, but I don't have enough rep $\endgroup$ – Alucard.cH Aug 11 '16 at 15:37
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    $\begingroup$ Unfortunately, it is not the right answer! $\endgroup$ – Hitesh Dholaria Aug 11 '16 at 15:48
  • $\begingroup$ @HiteshDholaria Ha-ha, ok, so the answer is 1 but still remains the question "why"? $\endgroup$ – rhsquared Aug 11 '16 at 15:50
  • $\begingroup$ @RadoslavHristov I don't know why people think only in binary, sounds like a common psychology error, though it could be a trap as well :) $\endgroup$ – Hitesh Dholaria Aug 11 '16 at 15:54
  • $\begingroup$ @HiteshDholaria Obviously it was meant as a joke. As the base is not mentioned it might be any digit from 0 up to F (assuming you are not using some exotic notations). $\endgroup$ – rhsquared Aug 11 '16 at 17:06

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