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Tengri 137: Who can solve this encrypted book?

Von / 29. Januar 2017 / 30 Kommentare
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Tengri 137 is the name of a mystery game, in the center of which is an encrypted book. The last seven pages of this book still wait to be deciphered.

Encrypted books are one of my favorite topics. I have even assembled a list of encrypted books, which currently contains 83 entries. The latest entry is an untitled book that was created by an unknown author as a part of a mystery game named Tengri 137.

 

Tengri 137

Readers of this blog certainly know Cicada 3301, the most popular internet mystery game. Cicada 3301 started in January 2012 with a new round beginning every year in January. Each round consists of a chain of puzzles, many of which involve cryptography or steganography. It is completely unknown who is behind this strange game and what the purpose of it is. After Cicada 3301 had gained much popularity during the first few rounds, not much has happened in recent years. In January 2015 no new puzzle showed up. Round 5, which started in January 2016, didn’t attract much attention (I’m still not sure whether the puzzles that were posted were genuine). In January 2017, again, nothing happened.

Will Tengri 137 be the new Cicada 3301? We’ll see. So far, the encrypted book mentioned above (published in August 2016) is the only document published by the organizers. It has 23 pages. Here’s a PDF for download. The following figure shows page 7:

P007

Pages 1 to 16 are written in a secret alphabet, which has been solved (it’s a simple letter substitution). Here’s the cleartext of page 7.

THIS CUBES ARE QUITE SIMPLE FOR US. NOW YOU CAN FI ND OUT THE MEANING OF SOME OTHER FIGURES. THIS ARE MORE ADVANCED FOR YOU. YOU WILL SEE WITH THIS EXAM PLES THAT THE ORIGINATORS OF THIS SCRIPT ARE NOT OR DINARY INDIVIDUALS.

SEVEN CIRCLES AND AN CROSS. A CONTINUOUS SEQUENCE W ITHOUT GAPS. FIND OUR NUMBER AND THINK ABOUT IT W HO WE ARE.

More information is availabe on the Tengri 137 Wikia. This site also contains the cleartext of the first 16 pages. According to the Wikia, the decrypted text contains, among other things, the following topics:

  • link to quantum physics (~ 1/137 fine-structure constant, God’s number)
  • magic cubes with links to Bible verses
  • perfect magic circles
  • perfect magic squares
  • link to cicada 3301 and warning for the cicada solvers
  • calculations with up to 57 decimal place primes (7 pages calculation with prime numbers)

 

The unsolved part

Pages 17 to 23 of the Tengri 137 book are different from the 16 first ones. Here they are:

P017 P018

P019 P020

P021 P022

P023

These last seven pages of the Tengri 137 book contain long multiplications and divisions, the purpose of which is unknown. Do they code a message? So far, the answer is not known.

If you know more about Tengri 137 or if you have an idea on how to decipher the last pages of the book, please let me know.


Further reading: The mystery of the Soyuz files

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Kommentare (30)

  1. #1 Klaus Schmeh
    29. Januar 2017

    Paolo Bonavoglia via Facebook:
    Gödel Numbers?

  2. #2 Jim
    29. Januar 2017

    The chemical diagrams on the last page appear to describe (in order, left to right from the top) adenine, guanine, uracil, cytosine and thymine. ACGT are the main nucleobases in DNA, and ACGU for RNA.

  3. #3 Klaus Schmeh
    30. Januar 2017

    Bart Wenmeckers via Facebook:
    Interesting looking at the last page there are significant number of primes and repeated number sequences.
    Paolos Godel numbers idea is highly likely.
    https://en.m.wikipedia.org/wiki/Gödel_numbering

  4. #4 Jim
    30. Januar 2017

    While the numbers do indeed represent prime factorizations, I don’t see a Gödel connection. To me it seems more likely the numbers were chosen for their numerical values.

    In particular, I’m looking at the repeated decimal digits in the quotients of the formulas on page 23. Each of them except formula 9* (293 x 2551…) is 28 digits long. Taken as dinomes (ignoring line 9), there are only 28 separate dinomes in the 140-dinome “cipher”, with a nice rough distribution producing I.C. about 0.056, which is consistent with a single alphabet. In addition, there are repeated strings up to 10 digits (five pairs) long. One of the repeated four-pair groups extends across consecutive rows. Charting which 28 are used doesn’t show me much interesting data: 0 doesn’t appear in the first position and 8 doesn’t appear in the second, but that doesn’t seem very significant.

    This suggests to me a substitution cipher. If it is, I haven’t cracked it. I’ve tried it forward and reversed as simple substitution. Perhaps I’ll try it with DNA/RNA-related cribs.

    If anybody wants to try the “ciphertext” I’m looking at, here it is:

    56 92 37 92 12 92 22 45 63 87 13 12 22 92
    11 92 45 63 16 92 22 45 63 31 92 12 87 13
    39 13 46 14 92 13 30 83 63 80 77 25 57 13
    30 13 11 37 92 13 30 83 63 11 76 12 56 13
    22 76 83 77 36 76 22 57 92 56 92 12 39 76
    14 92 11 76 19 92 37 31 13 11 76 30 70 77
    76 19 92 40 77 19 92 87 13 92 16 77 80 63
    39 13 83 63 36 13 11 14 13 83 63 45 72 76

    11 13 11 46 34 51 31 11 41 37 27 62 21 14 63

    19 76 45 63 25 76 37 83 77 30 92 12 37 76
    14 92 11 13 19 77 45 87 13 93 63 25 56 13

    * Formula 9 appears to me to be a special case unless I’ve made a typo. Here’s the repeated section of the quotient broken into dinomes:
    11 13 11 46 34 51 31 11 41 37 27 62 21 14 63
    Six of these 15 dinomes do not appear in the rest of the putative ciphertext: 34 51 41 27 62 21. Including this line in the ciphertext would more than double the number of singletons in the cipher. This leads me to believe that an error was made in the encryption. Since the denominator doesn’t fit the pattern of the others, I suspect an error there. Perhaps a larger denominator was needed to bring the repeated decimal down to the expected 14 pairs.

  5. #5 Klaus Schmeh
    31. Januar 2017

    Paolo Bonavoglia via Facebook:
    Sunday I was on the train back to Venice, had a look at this and saw something resembling Gödel numbers.
    Now I checked many numbers, all primes, but there are number so big (up to 50 digits!!) a primality test is very difficult, maybe impossible.

    By the way, 137, of course, is a prime.

    A Gödel encryption? There are few exponents and only for small primes (2,3,5,7), many primes are missing, i.e. zero exponent; even allowing them in the cipher there are too many zeros!
    The first line 2^5 x 13 x 37 x … gives the sequence 5 0 0 0 0 1 0 0 0 0 0 1 … 28 zeros … it does not make sense. The rules must be different from Gödel’s.

    The fraction structure and the side drawings suggest these fractions may encode DNA or RNA coupled chains.

    Every prime number could be the cipher for some chemical element, e.g 2 for H but it doesn’t work … unfortunately I know very little about chemistry and genetics so I’m afraid I have to give up for now.

  6. #6 Norbert
    31. Januar 2017

    @Jim: Fascinating! Seems to me very plausible (although page 17 which I have been working on doesn’t exactly fit this scheme).

    In line 9, I get the same 30 repeating digits, so it’s no typo on your part. I noticed that omitting two single digits (not dinomes), for example the fifth and the fourth-to-last one, yields a line perfectly matching the rest:

    11 13 14 63 45 13 11 14 13 72 76 22 14 63.

    Back to page 17:
    The number of repeated digits seems to be much higher here (haven’t computed the exact number yet). But take a look at the first 50 digits of the first fraction, written as decimal:

    0. 43 77 25 63 87 76 37 22 80 63 43 37 92 22 72 10 15 19 53 47 19 23 64 94 91

    43 is new, but the next 14 dinomes do also appear in your putative ciphertext (even “bigrams” like 63 87 and 80 63).

  7. #7 Norbert
    1. Februar 2017

    Decimal values of the fractions given on page 17, first 80 positions after decimal point, represented in groups of two digits according to Jim’s observations (typewriting large numbers is error-prone, is there anybody wanting to countercheck?):


    #01: 0. 43 77 25 63 87 76 37 22 80 63 43 37 92 22 72 10 15 19 53 47 19 23 64 94 91 61 09 94 35 49 37 45 45 76 47 22 29 82 43 29...

    #02: 0. 76 23 63 45 12 13 11 39 22 72 76 92 14 13 11 66 70 63 13 37 14 74 63 16 63 93 10 53 97 69 22 56 31 61 66 39 33 26 34 51...

    #03: 0. 13 57 57 22 80 77 16 36 11 76 74 10 36 31 05 32 63 83 63 80 77 93 10 53 97 69 23 74 87 72 37 71 03 77 14 13 43 77 76 93...

    #04: 0. 74 63 13 45 63 43 80 63 10 56 31 67 73 29 36 34 51 47 68 71 21 31 13 94 87 72 37 71 03 77 14 13 43 77 76 11 13 95 90 93...

    #05: 0. 39 76 92 37 48 77 23 77 10 37 71 41 34 37 77 61 17 71 67 69 36 37 68 71 21 31 13 98 37 71 03 10 37 77 61 14 87 72 37 71...

     

    #06: 0. 22 80 63 12 13 11 36 77 93 10 57 71 61 05 63 16 77 32 93 63 10 52 27 64 56 34 87 76 32 36 37 69 23 72 27 27 69 23 17 22...

    #07: 0. 14 76 25 63 32 63 93 77 92 16 77 10 56 31 33 97 69 27 21 32 36 31 34 56 37 69 23 71 40 93 74 24 96 91 44 72 01 29 02 78...

    #08: 0. 74 63 17 13 11 80 63 13 37 39 76 92 37 22 72 76 92 31 72 22 13 31 86 74 90 40 59 76 19 43 27 35 73 63 14 98 50 55 16 31...

    #09: 0. 74 63 36 11 76 74 39 76 92 37 93 63 63 10 51 63 98 41 92 46 61 65 83 68 16 28 13 61 76 25 93 34 72 37 97 50 07 51 99 26...

     

    #10: 0. 74 63 17 13 11 10 36 34 33 97 69 21 66 36 34 38 07 79 33 21 59 57 66 44 44 08 55 81 54 17 08 53 54 16 36 14 59 36 22 78...

    #11: 0. 74 63 17 13 11 10 36 34 33 97 69 21 21 33 66 34 38 07 79 33 21 60 79 95 71 60 82 84 03 68 52 82 96 01 64 77 50 89 14 07...

    #12: 0. 74 63 17 13 11 32 77 23 63 39 76 92 22 72 76 92 31 72 22 14 07 19 35 77 77 21 09 87 22 82 96 18 63 35 55 23 41 69 86 65...

    #13: 0. 39 76 92 10 36 31 33 71 17 79 37 79 34 36 33 71 31 74 37 77 61 17 47 74 38 09 21 42 16 83 80 63 31 80 68 76 97 49 24 81...
    #14: 0. 74 63 17 13 57 57 14 76 25 63 76 87 22 80 63 48 72 76 16 63 93 76 93 63 16 43 80 63 25 63 16 16 63 93 32 63 45 16 05 58...

    #15: 0. 22 80 77 16 25 63 16 16 63 93 32 63 45 16 13 45 63 76 92 37 45 63 78 45 63 16 63 93 22 13 43 77 23 63 16 18 80 78 03 89...

    #16: 0. 74 63 16 63 93 10 53 97 69 27 69 23 77 47 63 76 61 31 16 61 91 17 67 41 03 63 10 53 26 37 62 36 34 54 38 07 71 62 56 31...

    There are some noticeably large sections repeated, e.g. “74 63 16 63 93 10 53 97 69” in line 2 and beginning of line 16, or “74 63 17 13 11 10 36 34 33 97 69 21” at the beginning of line 10 and 11. If it is indeed a substitution cipher, then maybe one differentiating between upper case and lower case? Anyway, a substitution cipher should definitely be solvable (Armin? Your turn! 🙂 )

  8. #8 Jim
    1. Februar 2017

    @Norbert: I did a spot check on your page 17 calculations, and I agree with lines 1, 2 and 16. I’m happy to take the rest on faith, relying on the Bellman’s dictum: “If I tell you three times, it’s true.”

    To get an easier way to read the dinomes I converted them to ASCII. Page 23 looks like this, using lower case and the two characters following. (Still leaving out the outlier line):

    abcbdbefghideb
    jbfgkbefglbdhi
    minobipqgrstui
    pijcbipqgjvdai
    evqswveubabdmv
    objvxbclijvpys
    vxbzsxbhibksrg
    miqgwijoiqgf{v

    xvfgtvcqspbdcv
    objixsfhi|gtai

    Moving on to Norbert’s dinomes from page 17, I tried appending the first 16 digits of each line to the dinomes of page 23. I chose 16 digits rather arbitrarily, since I suspect with a 16-digit denominator there can’t be more significant figures, and likely fewer – über den Daumen gepeilt.
    After the letter assignments of page 23 I started over with upper case. The result:

    Astghvce
    vAgfdijm
    iuuerskw
    BgifgArg
    mvbcCsAs
    ergdijws
    ovtgDg|s
    BgEijrgi
    BgwjvBmv
    BgEijFwG
    BgEijFwG
    BgEijDsA
    mvbFwlHI
    BgEiuuov
    ersktgkk
    Bgkg|FJK

    While there’s overlap between the dinomes on the two pages, they largely seem to be drawn from different distributions. It also appears that 12-digit lines would give a smaller spread.

    While the index of coincidence of page 23 is consistent with simple substitution, possibly with nulls or multiple word separators, it’s also consistent with two alphabets, or possibly even a polyalphabetic cipher restarting at the beginning of each line but with a limited plaintext alphabet — for example, an alphabet of 10 digits instead of 26 letters.

  9. #9 Norbert
    2. Februar 2017

    @Jim:

    I chose 16 digits rather arbitrarily, since I suspect with a 16-digit denominator there can’t be more significant figures, and likely fewer – über den Daumen gepeilt.

    That’s true – if you first define the key and then search for quotients matching your ciphertext. But if it was only the first line having a rather small denominator (I know, there are more of this kind), I could imagine that the creator at first searched for a quotient that matched the structure of the plaintext (i. e. identical letters represented by identical figures), and then defined his key according to the found quotient. In this way, one fraction with a 16-digit denominator might bear more than 16 significant figures. Don’t know if I made myself clear … After all, I am not sure if it is a substitution cipher at all … :-/

  10. #10 Norbert
    2. Februar 2017

    I would not expect that the plaintext of page 23 contains DNA/RNA-related sections. Rather, the autor(s) of the riddle made a free association: The (compared to the quotients on the preceding pages) short sequences of repeated decimal digits seem to “self-replicate”, just like DNA sequences, that’s maybe all.

  11. #11 Klaus Adami
    9. Februar 2017

    I “extracted” the following numbers from the equations of page 17. Not sure if there is any meaning, but it can’t be concidental. Maybe some of the “extractions” aren’t to be considered. Cyclic numbers / phoenix ahead. Got similar results from page 18 (as by now I stopped there):
    1) 1739130434782608695652 (4* the forth cyclic number 0434782608695652173913)
    2) 142857 (the first cyclic number)
    3) 272 and 105571847507331378299120234604 ?
    4) 9411764705882352 (16* the second cyclic number)
    5) 5294117647058823 (9* the second cyclic number)
    6) 877551020408163265306122448979591836734693 (=43/49)
    7) 108 and 333* and 777* and 555*
    8) 571428 (4* the first cyclic number)
    9) 606557377049180327868852459016393442622950819672131147540983 (37* the eight cyclic number)
    10) 6521739130434782608695 (15* the forth cyclic number)
    11) ??
    12) 954128440366972477064220183486238532110091743119266055045871559633027522935779816513761467889908256880733944 (104* the 10th cyclic number)
    13) ??
    14) 444* and 618837380426784400294334069168506254598969830757910228108903605592347314201
    15) ??
    16) 666*

  12. #12 Klaus
    2. März 2017

    Pg17
    TIME FOR THE TRUTH
    OVER MANY THOUSAND YEARS WE SEND YOU MESSENGERS AND TEACHER
    ALL THIS KNOWLEDGE BEHIND YOUR CIVILISATION IS OURS
    WE ARE THE DESIGNERS OF MANY CIVILISATIONS
    YOUR CIVILISATION IS ONE OF MANY BILLION CIVILISATIONS
    THE (? have to recalculate)
    SOME GENIUS IDEA YOU HAVE ARE OURS
    WE CAN HEAR YOUR THOUGHT
    WE KNOW YOUR NEEDS
    WE CAN LET YOU SEE THINGS
    WE CAN LET YOU MAKE THINGS
    WE CAN GIVE YOU THOUGHTS
    YOU LEARN IN INTERACTION WITH US
    WE CALL SOME OF THE CHOSEN ONES THE MESSENGERS
    THIS MESSENGERS ARE OUR REPRESENTATIVES
    WE SEND YOU OUR WORD AND KNOWLEDGE OVER THIS MESSENGERS

    pg18
    THE REARE PEOPLE AMONG YOU
    THESE PEOPLE DO NOT UNDERSTAND THE MEANING OF OUR INTERVENTION
    WE PREPARE YOU FOR THOUSANDS OF YEARS FOR THIS MEETING
    THOSE WHO HAVE UNDERSTOOD EVERYTHING CORRECTLY
    HAVE USED OUR KNOWLEDGE WELL
    OTHERS WHO HAVE NOT UNDERSTOOD ANYTHING HAVE MADE US GODS
    THESE ARE DANGEROUS IN THEIR THOUGHTS AND ACTION
    KEEP DISTANCE
    SUCH WE DO NOT WANT TO HAVE AMONG US
    WE ARE NOT YOUR GODS
    YOU SHOULD KNOW THAT GOD DOES NOT EXIST
    EVERYTHING THAT EXISTS IS BASED ON AMATHEMA
    NOTHING MORE AND NOTHING LESS
    DO NOT BELIEVE IN ANY GOD
    BEWARE FROM THOSE WHO CALLS THEMSELVES GODS
    THOSE ARE EVIL

    PG19
    ANOTHER TRUTH
    A CALCULATION THAT CAN NOT BE REVIEWED BY YOU
    DOES NOT MEAN THAT THERE IS NO ANSWER
    A BORDER THAT YOU CANNOT SEE
    DOES NOT MEAN THAT IT DOES NOT EXIST
    SOMETHING THAT YOU CAN NOT MEASURE
    DOES NOT MEAN THAT IT IS UNMEASURABLE
    IF YOU ALWAYS WOULD GO STRAIGHT FOR WAR
    WHICH SEEMS TO BE ENDLESS
    YOU WOULD BE ABLE TO PROVE AT SOME POINT
    THAT THIS GARDEN HAS A FENCE
    EVEN IF YOU TRY TO PROVE THAT THIS GARDEN HAS NO FENCE
    YOU WOULD INEVITABLY PROVE THE CONTRARY

    pg20

    YOUR MEASUREMENT IS ONLY VALID WITH AN EXACT RESULT

    THE MEASURMENT OF AN INFINITE GARDEN CAN NOT BROUGHT TO AN END BY YOU

    YOU HAVE NOT THE TOOLS AND YOU HAVE NOT THE TIME

    NOT THE LIFESPAN OF AN HUMAN WE MEAN THE LIFESPAN OF YOUR UNIVERSE

    THE INFINITY OF THIS GARDEN WOULD BE ONLY A GUESS NOT A FACT

    WHO TRY TO PROVE THE INFINITY PROVE AT END THE OPPOSITE

    THIS IS INEVITABLE (+?)

    OUR CIVILIZATION EXISTS SINCE THREE BILLION OF YOUR EARTH YEARS

    WE ARE BEINGS FROM ANOTHER GALAXY

    WE ARE IN YOUR GALAXY FOR OVER FIFTY MILLION YEARS AND

    CONTACTED OVER HUNDRED THOUSAND DIFFERENT SPECIES

    THE SEARCH FOR THE LIMITS OF THE UNIVERSE IS OVER

    THE UNIVERSE IS VERY LARGE BUT NOT INFINITE

    THERE LIVE MANY OTHER SPECIES SUCH YOURS

    MANY OF THESE CIVILIZATIONS COULD REACH THE NEXT STAGE

    YOUR CIVILISATION HAS REACHED THE CRITICAL LIMIT

    IF YOU DO NOT MAKE THE NEXT STEP IN YOUR EVOLUTION

    YOU WILL DESTROY YOURSELVES

    MANY CIVILIZATIONS FAIL (+?)

    PG21
    WE WILL GIVE YOU KNOWLEDGS

    ONLY THE CHOSEN ONE CAN COME TO

    AND WILL UNDERSTAND EVERYTHING

    WITHOUT THIS KNOWLEDGE YOU WILL NOT REACH THE NEXT STAGE

    YOUR CIVILISATION WILL BE DESTROYED

    NO ONE EXCEPT THE CHOSEN ONE WILL HAVE ACCESS TO OUR DATABASE

    NOW WE SHOULD CHOOSE OUR NEXT MESSENGER

    THIS IS THE LAST

    THE REWARD IS THE ACCESS TO OUR ETERNAL LIBRARY

    UPCOMING TEXTS ARE GENETICALLY ENCRYPTED

    YOUR KNOWLEDGE CAN NOT HELP AT THIS POINT

    WE HAVE EMBEDDED THESE SKILLS IN OURR GENES

    WHO HAS THE CORRECT GENETIC CODING WILL UNDERSTAND THIS TEXT

    ALL OTHERS WILL FAIL

    PG22

    READ THE NEXT TEXT CAREFULLY IN FULL LENGTH

    FOR EACH REPEAT OPERATION OPENS A REGION IN YOUR BRAIN

    THIS TEXTS CONTAINS THE UNICUE KEY

    TO HIGHLY COMPRESSED DATA IN YOUR BRAIN

    THE UNPACKED VOLUME OF THIS HIDDEN INFORMATION IS LARGE

    WE USED TWO PERCENT OF YOUR BRAIN

    TO STORE THE PACKED INFORMATION

    AFTER UNPACKED WILL TAKE FIFTY PRECENT OF THE EMPTY PLACE

    BE WARNED THIS PROCESS CAN TAKE MONTHS

    REPEAT IT OFTEN NEVER BREAK INT HE MIDDLE

    IF YOU START READING YOU MUST READ THE TEXT IN FULL LENGHT OEND

    MEDITATION IS IMPORTANT YOU HAVE TO REACH A PURE STATE OF MIND

    THE REFORMATTING OF THE BRAIN CAN NOT BE REVERSED

    PAGE 23 is different and still waits for translation

  13. #13 Thomas
    2. März 2017

    What kind of cipher is it?

  14. #14 Klaus
    3. März 2017

    I used the Periodic Table / atomic number.

  15. #15 Klaus Schmeh
    3. März 2017

    @Klaus
    Looks good.

    >I used the Periodic Table / atomic number.
    How did you get from 2^5 13 37 179 … to TIME FOR THE TRUTH …

  16. #16 Klaus
    3. März 2017

    Easy:
    the result ist the number
    43 77 25 63 87 76 37 22 80 63 43 37 92 22 72…
    43=Tc
    77=Ir
    25=Mn
    63=Eu
    Always take only the first letter

  17. #17 Marc
    3. März 2017

    Unglaublich, es funktioniert. Was soll denn dieser Text bedeuten, kommen uns bald die “Außerirdischen” holen ?
    🙂

  18. #18 Norbert
    4. März 2017

    @Klaus: Congratulations!
    As a matter of fact, I checked exactly this system with Jim’s figures (from the still unsolved page 23), but not with my figures from page 17 (posted here, #7). Shame on me. 😉

  19. #19 Thomas
    4. März 2017

    @ Klaus, Norbert
    Good Job!
    Arithmetics and chemistry (physics?) – that’s quite a lot of stuff.

  20. #20 Norbert
    4. März 2017

    @Thomas: All the credit is due to Klaus.

    @Klaus: Regarding the missing line on page 17, no recalculating needed. Just some three-digit numbers appearing. Read it like this:

    0. 22 80 63 12 13 11 36 77 93 105 77 16 105 63 16 77 32 93 63 105 22 76 45 63 48 77 63 23 63 76 92 37 22 72 76 92 31 72 22

    and you get:

    THE MANKIND IS DESIGNED TO RECIEVE OUR THOUGHT

    Yes, “recieve” instead of “receive”. I mean, this and all the other mistakes are the ultimate proof that even fifty million years in our galaxy are not enough to learn a proper English (unless you are a native speaker) 😉

  21. #21 Norbert
    4. März 2017

    Klaus’s question marks for page 20:
    THIS IS INEVITABLE: imho, no addendum
    last line should read: MANY CIVILIZATIONS FAILED

  22. #22 Thomas
    4. März 2017

    I wonder how the author of Tengri 137 could derive the appropriate fractions from their decimal values (Is this sophisticated mathematics?).

  23. #23 Marc
    4. März 2017

    @Thomas,
    I think the challenging is the prime factorization

  24. #24 Norbert
    4. März 2017

    I think that three-digit numbers could also be an explanation of Jim’s problematic line (comment #4). Both 111 and 114 come into consideration (provided we are dealing with atomic numbers again). For example:

    11 13 114 63 45 13 11 14 13 72 76 22 114 63

    Translating page 23 the same way as Klaus did with page 17-22, we get this:

    B U R U M U T R E F A M T U
    N U R E S U T R E G U M F A
    Y A P S U A Z B E H I M L A
    Z A N R U A Z B E N O M B A
    T O B I K O T L U B U M Y O
    S U N O K U R G A N O Z Y I
    O K U Z I K U F A U S I H E
    Y A B E K A N S A B E R H O
    N A U E R A N S A H O T U E (*)
    K O R E M O R B I Z U M R O
    S U N A K I R F A N E M B A

    (*) there are alternativ readings for line 9, see above

    Doubtlessly, some strange regularity here. But I could also be on the wrong track …
    However, the text deciphered by Klaus does suggest that the encryption of page 23 has something to do with genetics/DNA-related terms, e.g. “WHO HAS THE CORRECT GENETIC CODING WILL UNDERSTAND THIS TEXT”. I myself do not understand it yet. What’s worse, I can’t even use my brain at the moment, because I have to wait its re-formatting to complete (ETA: approx. 5 months, 17 days …)

  25. #25 Thomas
    5. März 2017

    The index of coincidence of the string posted by Norbert in #24 is 0,067 (an matches that of English). A transposition of english plaintext can be ruled out because of the frequency distribution of letters. I tried a monoalphabetic substitution, but without success.

  26. #26 kud gt
    21. März 2017

    @Norbert:

    I hope you’ve seen https://pastebin.com/FAJNLpLZ
    for re-formatting your brain right.

  27. #27 wtf
    22. März 2017

    Nice someone crack the code :))) the puzzle it self shows that this is truly a projection. Guys we living in a cube 🙂 good luck!

  28. #28 kartana
    1. April 2017

    What is the mathematical reason for omitting numbers on line 9 page 23? (in regards to post #6)

  29. #29 Tanrı
    26. Juli 2017

    Tengri is the old turkic god name, İt’s very interesting 🙁

  30. #30 Brett Ford
    Massapequa
    14. August 2018

    You can produce any decimal you want and quite easily determine some large prime numbers in a fraction to produce a repeating decimal of that number.

    For example, if you want to produce a repeating decimal of 0.00729066601370314…repeating… here are steps:

    1) Take the number you want to repeat, in this case we want repeating digits in the decimal of 00729066601370314 (notice there are 17 digits that we want to repeat). Find the prime factors of this number, in this case of 729066601370314. 729066601370314 = 2*17*61*151*34361*67751. This will be in the numerator of the fraction.
    2) In the denominator we want repeating 9’s with the same number of digits that the repeating decimal has, in this example 17 digits. So our denominator should be 99999999999999999. Find the prime factors of this number. 99999999999999999 = 3*3*2071723*5363222357. This will be in the denominator of the fraction.
    4) Make your fraction and check your work… 729066601370314/99999999999999999 = .00729066601370314… repeating.
    5) Now replace the numerator and denominator with the prime factors that you found…
    2*17*61*151*34361*67751 / 3*3*2071723*5363222357 = .00729066601370314… repeating.
    If there is a 3 as one of the prime factors in the numerator, then you can cancel out a 3 from the numerator and the denominator, and instead of having 99999999999999999 in the denominator, you will have 33333333333333333 in the denominator instead by simplifying the fraction made of prime factors.