wolfgang Posted December 8, 2011 Report Share Posted December 8, 2011 Out of these combinations,find out number 34...have a good time.... 1d,4b,4c,1a,3a,2c,2b,3d,2a,3c,3b,2d,4d,1b,1c,4a Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 8, 2011 Report Share Posted December 8, 2011 I just had this idea... I don't know if this is the answer you are looking for... If we assume the base 24, 0-9 numbers and letters a onwards from 10-23 then 1a = 34 Similarly, in base 23, 1b=34 In base 22, 1c=34 In base 21, 1d=34 Quote Link to comment Share on other sites More sharing options...
0 wolfgang Posted December 8, 2011 Author Report Share Posted December 8, 2011 I just had this idea... I don't know if this is the answer you are looking for... If we assume the base 24, 0-9 numbers and letters a onwards from 10-23 then 1a = 34 Similarly, in base 23, 1b=34 In base 22, 1c=34 In base 21, 1d=34 nope...sorry....try something else...ty Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 9, 2011 Report Share Posted December 9, 2011 How about 3d? 3 is the tens digit, and d = 4 if you go by 1 = a, 2 = b, and so on. thus, 3d = 34 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 9, 2011 Report Share Posted December 9, 2011 34 is in place or equal to 3d.............. Quote Link to comment Share on other sites More sharing options...
0 mewminator Posted December 9, 2011 Report Share Posted December 9, 2011 (edited) are the combinations arranged, or are they random, or am I to figure that out? let the number be x and the letter y x could be tens and y units (in that case the answer is 3d) or vice versa (4c) x could be the number of my current cycle around the alphabet and y could be an addition to it (26 (x - 1) + y) but that won't work unless I tweak it like this (26 (x - 1) + xy) then the answer would be 2d one of a, b, c, or d = 17 or 34 and the answer is 2a, 2b, 2c, 2d, 1a, 1b, 1c, or 1d unless I decide to include fractions then it would just be a mess (11.333 or 8.5) x could be how many times y (as a letter) has been included in the word thirty-four... the only problem is that 34 doesn't contain a's, b's, c's, or d's multiplying x by y (as a number) then adding it to the next number until I reach 34: 1d + 4b + 4c +1a +3a +2c =1 x 4 + 4 x 2 + 4 x 3 + 1 x 1 + 3 x 1 + 2 x 3 =4 + 8 + 12 + 1 + 3 + 6 = 34 therefore the answer is 2c I wonder if any of this is correct... or am I at least on the right track Edited December 9, 2011 by mewminator Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 9, 2011 Report Share Posted December 9, 2011 3d Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted December 9, 2011 Report Share Posted December 9, 2011 If you plot these as locations in a 4x4 grid, with rows labelled 1-4 and columns labelled a-d, the sequence Wolfgang has produced is symmetric about the center of the grid. That is, if you view the two characters as RC (row, column), and number the sequence items from 0 to 15, C(i) = 4-C(15-i), R(i) = 4-R(15-i) I think it's interesting that his sequence completely fills the 4x4 grid and extends no farther. I think it's interesting that his sequence exhibits this symmetry. I have NO CLUE how this relates to the location of 34. Sorry... Quote Link to comment Share on other sites More sharing options...
0 wolfgang Posted December 10, 2011 Author Report Share Posted December 10, 2011 (edited) are the combinations arranged, or are they random, or am I to figure that out? let the number be x and the letter y x could be tens and y units (in that case the answer is 3d) or vice versa (4c) x could be the number of my current cycle around the alphabet and y could be an addition to it (26 (x - 1) + y) but that won't work unless I tweak it like this (26 (x - 1) + xy) then the answer would be 2d one of a, b, c, or d = 17 or 34 and the answer is 2a, 2b, 2c, 2d, 1a, 1b, 1c, or 1d unless I decide to include fractions then it would just be a mess (11.333 or 8.5) x could be how many times y (as a letter) has been included in the word thirty-four... the only problem is that 34 doesn't contain a's, b's, c's, or d's multiplying x by y (as a number) then adding it to the next number until I reach 34: 1d + 4b + 4c +1a +3a +2c =1 x 4 + 4 x 2 + 4 x 3 + 1 x 1 + 3 x 1 + 2 x 3 =4 + 8 + 12 + 1 + 3 + 6 = 34 therefore the answer is 2c I wonder if any of this is correct... or am I at least on the right track These combinations are arrenged in such a manner so that any change in the sequence will not give the same results. Edited December 10, 2011 by wolfgang Quote Link to comment Share on other sites More sharing options...
0 wolfgang Posted December 10, 2011 Author Report Share Posted December 10, 2011 If you plot these as locations in a 4x4 grid, with rows labelled 1-4 and columns labelled a-d, the sequence Wolfgang has produced is symmetric about the center of the grid. That is, if you view the two characters as RC (row, column), and number the sequence items from 0 to 15, C(i) = 4-C(15-i), R(i) = 4-R(15-i) I think it's interesting that his sequence completely fills the 4x4 grid and extends no farther. I think it's interesting that his sequence exhibits this symmetry. I have NO CLUE how this relates to the location of 34. Sorry... Woooow you are very close....!! Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted December 10, 2011 Report Share Posted December 10, 2011 I sure don't FEEL close! By the way, there's a LOT more symmetry--the sequence appears to be a Hamiltonian circuit around a 2x2x2x2 hypercube. But I'm still missing a couple of fundamentals. 1) Perhaps we are to add one or two dimensions (egad, maybe 4 more dimensions!) 2) But his sequence always moves to a new dimension before finishing the old ones. However, this sequence has now completely filled the 2x2x2x2 without starting into new ones. 3) and I have no idea how 1-4 relate to a-d. If we see them as hexadecimal numbers, the gap between 4 and A is not a convenient one, I could be more comfortable (still clueless, but more comfortable) if a-d had been c-f. This is very interesting wolfgang, thanks for the workout...still working... Quote Link to comment Share on other sites More sharing options...
0 wolfgang Posted December 11, 2011 Author Report Share Posted December 11, 2011 I sure don't FEEL close! By the way, there's a LOT more symmetry--the sequence appears to be a Hamiltonian circuit around a 2x2x2x2 hypercube. But I'm still missing a couple of fundamentals. 1) Perhaps we are to add one or two dimensions (egad, maybe 4 more dimensions!) 2) But his sequence always moves to a new dimension before finishing the old ones. However, this sequence has now completely filled the 2x2x2x2 without starting into new ones. 3) and I have no idea how 1-4 relate to a-d. If we see them as hexadecimal numbers, the gap between 4 and A is not a convenient one, I could be more comfortable (still clueless, but more comfortable) if a-d had been c-f. This is very interesting wolfgang, thanks for the workout...still working... Thank you Dear...I am so glad to hear this....sometimes,the situation is not so complex as we immagine!...take it easier....!!! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 11, 2011 Report Share Posted December 11, 2011 Sum all the series from 1a to 4d, and divide them to 8, there is your 34 Quote Link to comment Share on other sites More sharing options...
0 wolfgang Posted December 12, 2011 Author Report Share Posted December 12, 2011 Sum all the series from 1a to 4d, and divide them to 8, there is your 34 ?? ..what?? Quote Link to comment Share on other sites More sharing options...
0 wolfgang Posted December 12, 2011 Author Report Share Posted December 12, 2011 we can obtain ten times 34 out of this sequence . Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted December 12, 2011 Report Share Posted December 12, 2011 (edited) Sum all the series from 1a to 4d, and divide them to 8, there is your 34 He thinks you gave the instructions to a magic square. 1d,4b,4c,1a,3a,2c,2b,3d,2a,3c,3b,2d,4d,1b,1c,4a 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 Which your new hint supports. Edited December 12, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 12, 2011 Report Share Posted December 12, 2011 He thinks you gave the instructions to a magic square. 1d,4b,4c,1a,3a,2c,2b,3d,2a,3c,3b,2d,4d,1b,1c,4a 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 Which your new hint supports. so, the 34 we've been looking for is everywhere...in every horizontal, vertical and main diagonal line. Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted December 12, 2011 Report Share Posted December 12, 2011 Yes, that solution is shockingly direct, curr3nt! Good job! What a relief! Quote Link to comment Share on other sites More sharing options...
0 mewminator Posted December 13, 2011 Report Share Posted December 13, 2011 (edited) I see a whole lot of 34's 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 I think that's enough Edited December 13, 2011 by mewminator Quote Link to comment Share on other sites More sharing options...
0 wolfgang Posted December 13, 2011 Author Report Share Posted December 13, 2011 He thinks you gave the instructions to a magic square. 1d,4b,4c,1a,3a,2c,2b,3d,2a,3c,3b,2d,4d,1b,1c,4a 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 Which your new hint supports. Yes...I gave the instructions to a magic square,by giving the squares on X-axes,letters(a,b,c,and d) and giving numbers(1 to 4) to Y-axes squares. now,1d will mean the square for 1, and 4b will indicate where number 2 should be,,,and so on Quote Link to comment Share on other sites More sharing options...
0 robo Posted December 14, 2011 Report Share Posted December 14, 2011 Using the grid mentioned by CaptainEd, I entered numbers 1-16 in the grid. 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 And I find that each row, column and diagonal sum up to 34 Quote Link to comment Share on other sites More sharing options...
0 wolfgang Posted December 14, 2011 Author Report Share Posted December 14, 2011 Using the grid mentioned by CaptainEd, I entered numbers 1-16 in the grid. 4 14 15 1 9 7 6 12 5 11 10 8 16 2 3 13 And I find that each row, column and diagonal sum up to 34 Thats right...and you can enter numbers in the grid according to my combination also. Quote Link to comment Share on other sites More sharing options...
Question
wolfgang
Out of these combinations,find out number 34...have a good time....
1d,4b,4c,1a,3a,2c,2b,3d,2a,3c,3b,2d,4d,1b,1c,4a
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