grey cells Posted June 28, 2008 Report Share Posted June 28, 2008 OK guys , take a break from solving riddles and concentrate a bit on math. 1 , 2 , 3 , 4 , 101 , 12 , 7 , 8 , .............. , 11001 , 62 , 33 , __ , __ .......... There are 16 terms missing between 8 and 11001 , then till infinity. So what comes next and next and ..... Quote Link to comment Share on other sites More sharing options...
0 unreality Posted June 28, 2008 Report Share Posted June 28, 2008 question is binary involved? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 28, 2008 Report Share Posted June 28, 2008 Every 5th term is in base 2 Every term after a base 2 is a base 3 etc etc. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 28, 2008 Report Share Posted June 28, 2008 Every 5th term is in base 2 Every term after a base 2 is a base 3 etc etc. I agree that every 5th term is in base 2 but every term after a base 2 is not a base 3 since in the sequence 1 , 2 , 3 , 4 , 101 , 12 , 7 , 8 , .............. , 11001 , 62 , 33 , __ , __ .......... the 62 would be in base 3 and you cannot have a 6 in a base 3 number. I am pretty certain that the second term after every base 2 term is in base 8. So 3, 7, and 33. Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted June 28, 2008 Author Report Share Posted June 28, 2008 you are , zzembower . You have a part of the reasoning . You are right , as in that , after base 2 , base 3 is not employed And Sega was correct , that is , the first part of her reasoning was spot on . I hope that answers your question , Unreality. EDIT : Further clarification. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 28, 2008 Report Share Posted June 28, 2008 The base 2 is right, but base 3, not so much. It seems that 12 is thrown in to confuse, or its double of the next base 1 number, not sure. However, if it was doubled, then it should be 52 not 62, so that 12 is just a loop. It counts up from 8 to 24 in between before jumping to base 2 again. Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted June 28, 2008 Author Report Share Posted June 28, 2008 The base 2 is right, but base 3, not so much. It seems that 12 is thrown in to confuse, or its double of the next base 1 number, not sure. However, if it was doubled, then it should be 52 not 62, so that 12 is just a loop. It counts up from 8 to 24 in between before jumping to base 2 again. Welcome to the den , McGraff . There are no distractions in this sequence . Just concentrate on the sequence and you will get it. And McGraff , please use spoilers as that will help the others to think in an unbiased way. I think someone will get the answer soon . I can see it around the corner. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 28, 2008 Report Share Posted June 28, 2008 28 then 11101 then 72 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 28, 2008 Report Share Posted June 28, 2008 (edited) I just found this Gadget. It seems to me we are talking four columns: Binary, Previous # plus 10, Octal, and Hex? So, the 16 missing terms would be: 1001, 22, 13, C, 1101, 32, 17, 10 10001, 42, 23, 14 10101, 52, 27, 18 followed by: 11001, 62, 33, 1C, etc, etc The Hex column could have been decimal (4, 8, 12, 16...) as I don't see a pattern from one column to the next. Edited June 29, 2008 by CipherHost Quote Link to comment Share on other sites More sharing options...
0 akaslickster Posted June 29, 2008 Report Share Posted June 29, 2008 could 101 count as 11? Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted June 29, 2008 Author Report Share Posted June 29, 2008 28 then 11101 then 72 Welcome to the den , nmsu. You almost got it , i.e. , except for 28. And please use spoilers in the future. Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted June 29, 2008 Author Report Share Posted June 29, 2008 ]I just found this Gadget. It seems to me we are talking four columns: Binary, Previous # plus 10, Octal, and Hex? So, the 16 missing terms would be: 1001, 22, 13, C, 1101, 32, 17, 10 10001, 42, 23, 14 10101, 52, 27, 18 followed by: 11001, 62, 33, 1C, etc, etc The Hex column could have been decimal (4, 8, 12, 16...) as I don't see a pattern from one column to the next. Good work Cipher . You got all the answers and please use spoilers in the future. My logic was slightly different . What I had in mind was a cycle of bases : base-2 , base-4 , base-8 , base-16 which are all powers of 2 (2^1 , 2^2 , 2^3 , 2^4). And welcome to the den , Cipher. Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted June 29, 2008 Author Report Share Posted June 29, 2008 could 101 count as 11? Aka . No twisting of rules. 101 is 101. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 2, 2008 Report Share Posted July 2, 2008 omg i dont understand this... then again, im only 12: Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted July 2, 2008 Author Report Share Posted July 2, 2008 omg i dont understand this... then again, im only 12: Hey ! IcE(mAn) , welcome to the den ! No need to worry , you will learn a fair bit about the number systems , if you are around the den for a few more days . Quote Link to comment Share on other sites More sharing options...
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grey cells
OK guys , take a break from solving riddles and concentrate a bit on math.
1 , 2 , 3 , 4 , 101 , 12 , 7 , 8 , .............. , 11001 , 62 , 33 , __ , __ ..........
There are 16 terms missing between 8 and 11001 , then till infinity.
So what comes next and next and .....
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