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Comentarios

• pythoncodingveryfun 🤓24s · 2024-02-09T07:38:41.061Z

  nice!

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• Steve Johnson 🤓34s · 2024-02-13T23:20:44.693Z

  Very nice!

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• Jack M 🤓22s · 2024-02-24T00:14:29.813Z

  fun, thank you!

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• Philip Chow 🤓39s · 2024-02-24T02:54:21.215Z

  Soba and somen didn’t fit … great puzzle! Now I’m hungry

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• Brian Callahan 🤓24s · 2024-02-24T03:14:33.090Z

  

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• ForeverJung 🤓2:02 · 2024-02-24T12:39:48.779Z

  Good challenge!

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• Amandel 🤓1:23 · 2024-02-24T14:06:38.267Z

  Loved it, thanks!

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• bruhbruhbr 🤓59s · 2024-02-24T17:23:51.673Z

  heck yeah

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• kurtalert 🤓20s · 2024-02-24T17:48:46.528Z

  Thank you!

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• Adam 🤓11s · 2024-02-24T18:48:18.378Z

  Technically, the clue for 4D is maximally wrong -- any sequence of digits is found an arbitrary number of times in an irrational number (I'm a 6D, I know)

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  • quatrevingtneuf constructor · 2024-02-26T06:46:14.034Z

    not necessarily! this is true of normal numbers, not irrational numbers in general. out-6D-ed

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    • Adam 🤓11s · 2024-02-26T16:05:10.749Z

      Oof :(

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    • Puzzled 🤓46s · 2024-02-27T17:12:34.393Z

      I'm still having trouble with that clue, though I enjoyed the puzzle regardless!

      Irrational numbers typically have all sorts of repetitions in them, but the clue is making a general statement about the opposite.

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      • quatrevingtneuf constructor · 2024-02-28T17:05:08.664Z

        while certain substrings may repeat, the expansion as a whole will never enter a repeating pattern

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        • Puzzled 🤓46s · 2024-02-29T00:11:12.532Z

          Ah, I get what you mean now. Thanks for clarifying.

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• Neathery 🤓52s · 2024-02-25T21:22:03.599Z

  Thanks!

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• James Brinkworth 🤓1:39 · 2024-02-25T22:34:53.484Z

  Neat!

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• firedude750 🤓1:03 · 2024-02-27T19:27:24.560Z

  Thanks

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• Keith VanEtten 3:25 · 2024-03-03T17:22:25.881Z

  Sweet!

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• BrettRKS 🤓48s · 2024-04-11T01:13:29.865Z

  Very nice

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