What answer is hidden below?

                     Translate: [59111519920519195181718113131201]


                   [16th]:   (5)(1 11)(1 11)(1)(6 32)(5)(1 4)(1 2)(1)
                   [21th]:  (2 19)(2 11)(19 2)(2 5)(19)(22)(2 3)(2 7)                     
  • $\begingroup$ I have about two dozen ideas about this (which might be enough for a partial answer later), but the (6 32) is really throwing me off as it doesn't seem to fit any of them. Could you please confirm there are no typoes? $\endgroup$
    – Bass
    Nov 15, 2023 at 8:11
  • 1
    $\begingroup$ @Bass Yes just checked, no typoes. $\endgroup$ Nov 15, 2023 at 8:26

1 Answer 1


This was quite a journey, with several "a-ha!" moments required to reach the conclusion. I left in some of the memo notes I wrote to myself while solving, hope it makes the thought processes a bit easier to follow. (In addition to hopelessly cluttering the answer up). In any case, here's what I got:

Starting with the "Translate" line, we can split it into numbers between 1 and 26, so we get (after some considerable guesswork and fiddling around with the connections of the various ones):

 5 9 11 15 19 9 20 5 19 19 5 18 1 7 18 1 13 13 1 20 1
 E I K  O  S  I T  E  S  S E  R A G  R A  M  M A  T A

The A1Z26 interpretation doesn't seem to make too much sense, although it definitely looks like a natural language. (Because as long as it didn't, we'd continue tweaking the "is 113 one-thirteen or eleven-three?" options.)

Sounding it out it might actually be

Greek transcribed with the English alphabet

And indeed that turns out to be the case:

"ΕΙΚΟΣΙ ΤΕΣΣΕΡΑ ΓΡΑΜΜΑΤΑ" means "Twenty Four Letters" in Greek.

The "16th" and "21st" are then very likely references to the (topically never ending)

mathematical constants $\pi$ (3.14..) and $\phi$ (1.618..), which are respectively the 16th and 21st letters of the 24-letter Greek alphabet.

The square brackets kind of suggest indexing (at least to someone with a programming background), and the "after." might be an instruction to ignore the part before the decimal point. But how on earth would one index with two numbers in parens?

Maybe it's a simple substitution cipher? Sure looks like one, because of all the repeating elements:

              [16th]:   (5)(1 11)(1 11)(1)(6 32)(5)(1 4)(1 2)(1)
                         I     M     M  E     D  I    A    T  E

              [21th]:  (2 19)(2 11)(19 2)(2 5)(19)(22)(2 3)(2 7)     
                           L     U     N    C   H   B    O    X

Yeah, that could work. Doesn't seem to make a lot of sense though, and didn't help with the indexing either.

After verifying with OP that the 32 (in the middle of the first row) isn't a typo, we kinda know we're not dealing with any direct alphabetic ciphers, which don't reach numbers that big. But what else could they be? After a good while of staring at the screen and pondering the constants hinted at in the puzzle, I finally figured out why OP would put such a number in the puzzle:

32 is actually

the position of the first zero in the decimal expansion of pi!

Now it all makes sense: the two-number groups are, quite simply, just two digit numbers!

So let's replace each number with

that many-eth digit of pi (3.14159 26535 89793 23846 26433 83279 50288...), then read any pairs as a single number, finally translating to English with A1Z26:

 [16th]:   (5)(1 11)(1 11)(1)(6 32)(5)(1 4)(1 2)(1)
 [16th]:   (9)(1  8)(1  8)(1)(2  0)(9)(1 5)(1 4)(1)
            9    18    18  1    20  9   15   14  1
            I     R     R  A     T  I    O    N  A

And repeating for the other line but this time indexing

the golden ratio: 1.61803 39887 49894 84820 45868...

  [21th]:  (2 19)(2 11)(19 2)(2 5)(19)(22)(2 3)(2 7)
  [21th]:  (1  2)(1  4)(2  1)(1 3)( 2)( 5)(1 8)(1 9)
              12    14    21   13   2   5   18   19
               L     N     U    M   B   E    R    S

we get the final answer of


which, as the title suggests, are indeed "never ending" when written in decimal. (Or in any other integer base, for that matter.) Nice!

  • $\begingroup$ You got it :) And yes the "after." just meant "after decimal point". Well done! $\endgroup$ Nov 15, 2023 at 13:25
  • $\begingroup$ @Prim3numbah nice red herring on the immediate lunchbox? how did you even do that, haha $\endgroup$
    – justhalf
    Nov 15, 2023 at 14:39
  • $\begingroup$ @justhalf no need to do anything, really: the unicity distance for an English language substitution cipher is well over 20 letters; any message shorter than that is likely to have spurious solutions like the lunchbox one above. $\endgroup$
    – Bass
    Nov 15, 2023 at 15:18
  • $\begingroup$ @Bass Ah, I had assumed you substituted the numbers to letters by a common shift. Apparently it's arbitrary, then it's much more likely, I agree. $\endgroup$
    – justhalf
    Nov 15, 2023 at 15:22
  • $\begingroup$ @justhalf applying a common shift would be a bit trickier, seeing how some of the numbers are actually pairs 🙃 $\endgroup$
    – Bass
    Nov 15, 2023 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.