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On this quiz, 005227 encodes the name Abby and 716123863 to Taylor. The key has something to do with encoding each letter as base 3 (c=010) and then writing the numbers in a column and reading down. Even given the answer I'm not able to make sense of it. Can anyone shred more light on the encoding scheme?

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  • $\begingroup$ if c =010 a =000? $\endgroup$ Commented Dec 18, 2023 at 5:16
  • $\begingroup$ The source of this puzzle is a book titled "Return to Montague Island: More Mysteries and Logic Puzzles", correct? $\endgroup$ Commented Dec 18, 2023 at 8:16
  • $\begingroup$ A sample from this book available on Amazon includes the answer keys from the back of the book. The answer for Puzzle 5.2, Quiz 5 says, rot13("hfvat gur fnzr zrgubq hfrq gb rapelcg gur zrffntr va gur Zrzbenoyr Zbfnvp", fb V cerfhzr gur zrgubq vf npghnyyl tvira gb lbh fbzrjurer ryfr va gur obbx.) $\endgroup$ Commented Dec 18, 2023 at 16:51

1 Answer 1

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This cipher is a...

hybrid of Delastelle's Bifid and Trifid ciphers.

How does it work?

First, the plaintext name is converted to a string of digits using base 3.

T   A   Y   L   O   R
202 001 221 110 120 200

Next, this string of digits is split in half and the second half is written below the first.

202001221
110120200

This produces several columns, where each column has two digits. Read down each column to read a two-digit base-3 number, and convert that number to a single base-9 digit (by multiplying the first digit by 3 and adding the second digit)

2 0 2 0 0 1 2 2 1
1 1 0 1 2 0 2 0 0
-----------------
7 1 6 1 2 3 8 6 3

You are left with the encoded form of the name TAYLOR which is 716123863 It is unclear from the examples what should happen to a name with an odd number of letters, but I imagine that a space could be added to the end and encoded as 000 in base-3.

But how are you supposed to solve it?

Figuring this out without being familiar with Delastelle's ciphers will be a bit tricky. You would need the insight that the encoded digits are actually base-9 digits (because the digit 9 does not appear, and also because the encoded digits are 50% longer than the names they encode, which is exactly long enough to be accounted for by a conversion from base-27 to base-9), followed by the impulse to convert both the name and the encoded digits to base-3. From here you can compare the two strings of digits to find out that the same base-3 digits appear in each but in different orders, and then try to deduce how the base-3 digits had been rearranged. Figuring this out without a clue that base-3 is involved would be even more challenging. I'm not sure how much of what was shared here makes up the entire puzzle as it appeared in the book, so I don't know what the intended solution really was.

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  • $\begingroup$ Excellent explanation, thank you $\endgroup$ Commented Dec 19, 2023 at 15:13

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