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This is a cryptogram puzzle, with some riddle-style clues to help you solve it. When you find the thing that links all of the clues, that will let you deduce the keys to decipher the cryptogram.

Here are the clues:

  • Christmas day

  • cross bones

  • t

  • amphibian surmounts

  • hnadwrintig

and here is the ciphertext:

Fbocov quq vdbwrv'x lbzf mji uxp vr tfdan;
Uli sbed "Mnd Thsouf jt" bbw lkq nys megrr.
Oh lbjmp rbh xmtn; gcb Dvtim xjudgttb "Fb!
Mua Gsrrjexh bv!" faifbavp gfv galncr huu.

One of the Rumkin cipher tools can be used for decoding.

You'll need two keys to decode the cryptogram; both keys are numbers with their digits spelled out in words, for example the number 1234567890 would be represented like this: onetwothreefourfivesixseveneightninezero (just an example, not one of the keys). Decimal points are ignored and there are no thousands separators or anything else except for the digits zero to nine.

You might want to refer to this question and its answers:
What characteristics of a ciphertext can be indicators of a particular cipher?
and you will probably need to use reference materials. Feel free to use any resources at your disposal (i.e. using Google is ok).

The plaintext consists mostly of English dictionary words.

Your challenge is to work out the two keys (and the reasoning behind them) and the plaintext. Also the reasoning behind each of the clues.

In the spirit of teamwork, partial answers are welcome.

I'd also be interested in constructive feedback on the puzzle itself, so my question is also "how could this puzzle be improved?"


To make this puzzle slightly easier, I'm adding a checksum for each of the keys. That way you can tell if you have one of the keys right (even if you don't have the other key or you're not sure what cipher is in use). The checksums are:

c40c8102ca3c33705325a8dd614c49a2d29c82f3fce529528f2fe4001fac8a233b0d6916f1c604643a9c375ca3e02589ed5babf8c9e47d0629e145ec13ca8ef0

and

a077332a1720b83b9a1c26619f3179dcfff1ef205f09468996c579aedd5ed5fb285acfb02c085768d8fb1cf2bfea05b02e993d947a324e361254bd68824e16e3

They are SHA-512 (SHA2) hashes generated using https://quickhash.com/hash-sha512-online. You can test possible keys by pasting the possible key into 'input data' and seeing what comes up in 'output data' when you click the 'generate' button.

Given that you already have quite a lot of information about the keys, it might be feasible to attack the hashes to find the keys. This would be a legitimate way of solving the puzzle as far as I'm concerned. But you'll still be expected to explain why those keys have been chosen. :)

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  • 3
    $\begingroup$ P.S. Happy Christmas! $\endgroup$ – A E Dec 24 '14 at 19:45
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    $\begingroup$ @randal'thor, i.imgur.com/842YGIR.jpg $\endgroup$ – A E Dec 27 '14 at 17:51
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    $\begingroup$ @AE That should be the slogan for Puzzling.SE! $\endgroup$ – Rand al'Thor Dec 27 '14 at 17:56
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    $\begingroup$ @AE - I assume the dyslexic-type spelling is deliberate, and you're not drunk or something? ;-) $\endgroup$ – Rand al'Thor Dec 27 '14 at 18:16
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    $\begingroup$ @AE This has all been done in the wrong order! First the numbers, then the answer to the riddle, then the explanation of the answer from the clues. $\endgroup$ – Rand al'Thor Dec 30 '14 at 13:10
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This answer gives the two numbers, got by brute-forcing the hashes rather than solving the riddle, and the final solution. For the solution of the riddle, see Len's and rand al'thor's answers.

I have cracked both hashes using the Mathematica program below:

Hash 1 is:

Hash 1 matched number 'sixsixseventhreeeightfour' (667384) It seems to be ASCII for "BIT".

Hash 2 is:

Hash 2 matched number 'nineeightzerosixsixfive' (980665) I can't seem to decipher the meaning, since one of the values isn't a print ASCII character.

The answer decrypts to the message:

Nature and nature's laws lay hid in night; God said "Let Newton be" and all was light. It could not last; the Devil shouting "Ho! Let Einstein be!" restored the status quo.

Using the Mathematica program

Do[
  iStr = ToString[i];
  iLen = StringLength[iStr];
  numStr = "";
  Do[
   posStr = StringTake[iStr, {j}];
   If[posStr == "0", numStr = numStr <> "zero"];
   If[posStr == "1", numStr = numStr <> "one"];
   If[posStr == "2", numStr = numStr <> "two"];
   If[posStr == "3", numStr = numStr <> "three"];
   If[posStr == "4", numStr = numStr <> "four"];
   If[posStr == "5", numStr = numStr <> "five"];
   If[posStr == "6", numStr = numStr <> "six"];
   If[posStr == "7", numStr = numStr <> "seven"];
   If[posStr == "8", numStr = numStr <> "eight"];
   If[posStr == "9", numStr = numStr <> "nine"];
   , {j, iLen}];
  iHash = IntegerString[Hash[numStr, "SHA512"], 16, 128];
  If[iHash == 
    "a077332a1720b83b9a1c26619f3179dcfff1ef205f09468996c579aedd5ed5fb2\
85acfb02c085768d8fb1cf2bfea05b02e993d947a324e361254bd68824e16e3",
   Print["Hash 2 matched number '" <> numStr <> "'"];
   ,];
  If[iHash == 
    "c40c8102ca3c33705325a8dd614c49a2d29c82f3fce529528f2fe4001fac8a233b0d69\
16f1c604643a9c375ca3e02589ed5babf8c9e47d0629e145ec13ca8ef0",
   Print["Hash 1 matched number '" <> numStr <> "'"];
   ,];
  If[Mod[i, 10000] == 0, Print["Tested up to " <> iStr];];
  , {i, 10000000}];
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  • 1
    $\begingroup$ Note: I will stop the code at 10M if nothing is found. Other more competent programmers are welcome to write faster implementations in lower-level languages. $\endgroup$ – March Ho Dec 28 '14 at 16:30
  • $\begingroup$ Good old Mathematica. ;) Your method looks sound to me. $\endgroup$ – A E Dec 28 '14 at 16:34
  • $\begingroup$ Ah, well done, you have everything except the five clues. How are they linked to the keys? $\endgroup$ – A E Dec 28 '14 at 16:43
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    $\begingroup$ Maybe 980665 is @AE's phone number? ;-) Seriously though, that number does sound familiar. $\endgroup$ – Rand al'Thor Dec 28 '14 at 19:01
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    $\begingroup$ A bit of googling suggests that 980665 is the reciprocal of a pascal to an atmosphere, using the standard definition of atmospheres. en.wikipedia.org/wiki/Pascal_%28unit%29 :P $\endgroup$ – March Ho Dec 28 '14 at 19:17
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+50
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This answer explains the significance of the two numbers found in MarchHo's answer, and also answers the riddle, but not in the way intended by the OP (for which see rand al'thor's answer).

Continuing with the keys, thanks to @MarchHo:

6.67384x10^-11 N m2/kg2 is G, the gravitational constant
9.80665 m/s2 is g, the standard gravity on earth

Possible meanings of the five clues are:

Christmas Day - Sir Isaac Newton was born on Dec 25, 1642 (Julian calendar)
cross bones - was Sir Isaac Newton's personal coat of arms
t - could be associated with the Samsung Gravity Touch phone
amphibian surmounts - could be a reference to "The Tale of Mr Jeremy Fisher". Jeremy Fisher is a frog who has a friend called Sir Isaac Newton
amphibian surmounts - could also describe overcoming gravity in adapting to dry land
hnadwrintig - could suggest that Sir Isaac Newton may have been dyslexic

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  • 1
    $\begingroup$ Hi Len - yes, that's the keys correct. You're also right on 'christmas day' and 'cross bones'. The other clues are mostly wordplay, like a cryptic crossword. $\endgroup$ – A E Dec 30 '14 at 10:06
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    $\begingroup$ Well done Len! I knew I'd seen 980665 before. @AE getting 'gravity' from those clues would have been pretty damn hard IMO. $\endgroup$ – Rand al'Thor Dec 30 '14 at 10:34
  • $\begingroup$ @rand not 'gravity' but rather... (What's a kind of amphibian? What does the symbol t stand for? What's an Apple product that appeared in Doonesbury?) $\endgroup$ – A E Dec 30 '14 at 11:07
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    $\begingroup$ @rand, that's right about the amphibian. :) Re the Apple product, we could say you've recognised 'handwriting' -> 'handwriting recognition'. I nearly made egg freckles a clue, it was a thing at the time. $\endgroup$ – A E Dec 30 '14 at 12:35
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    $\begingroup$ @randal'thor, the metric ton ('t') isn't an 'old ton', it's a ... $\endgroup$ – A E Dec 30 '14 at 12:47
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This answer details the first part of the solution of the riddle and says which cipher to use. For the two keys, see Len's answer. For the final solution, see MarchHo's answer.

As far as I can see, the only one of the Rumkin cipher tools that needs two keys is

the Keyed Vigenere cipher, which needs both an 'alphabet key' and a 'passphrase'.

Presumably (although he didn't say) this is what MarchHo used to decrypt the message.

The answer to the riddle is

NEWTON

which gives the two numbers found by Len. All the clues point towards this answer as follows.

Christmas day

Isaac Newton was born on Christmas Day 1642 [ref]

cross bones

were Isaac Newton's personal sigil [ref]

t

refers to the metric tonne or new ton (brilliant!)

amphibian surmounts

A newt is a kind of amphibian, and 'surmounts' is a fancy word for 'on', so this gives newt on (also brilliant!)

hnadwrintig

You recognised this word, didn't you? You realised it meant 'handwriting' even though that's not what it says. This is supposed to point to handwriting recognition and a device called Newton.

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  • $\begingroup$ @AE I'm now on to a new idea. Any joy? $\endgroup$ – Rand al'Thor Dec 26 '14 at 23:07
  • $\begingroup$ Still fairly chilly I'm afraid. $\endgroup$ – A E Dec 27 '14 at 8:28
  • $\begingroup$ I've added another clue $\endgroup$ – A E Dec 27 '14 at 10:34
  • 1
    $\begingroup$ @AE I've added another guess $\endgroup$ – Rand al'Thor Dec 27 '14 at 18:19

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