These words have a remarkable form of self-symmetry:






What is it?


It is astounding that @GarethMcCaughan solved this puzzle with so little information. I was fully prepared to drop successive hints for days until someone eventually solved it!

I was not only curious to see if anyone could solve it, but I was especially interested to see how they would explain the answer to a general audience. Again, I think Gareth's explanation is as clear and efficient as can be.

Here is more discussion (as well as a cool visual) for those who are not mathematically inclined:

I was exploring what happens when you take every other letter in a word, every third letter, every fourth letter, every fifth letter, etc. Of course, you quickly reach the end of the word, so in order to keep it fun and use all the letters, you have to wrap around to the beginning again. Like Pac-Man.

I wondered, when you take every third letter, or fourth letter, or fifth letter, do you ever get a new word out of it? It turns out, sometimes you do. For example:
— Starting from the T in the word THREADS and taking every second letter (remembering to wrap around when you reach the end) gives you TRASHED
— Starting from the T in the word TRASHED and taking every fourth letter gives you THREADS
— Starting from the P in the word SPRITES and taking every fourth letter you PERSIST
— Starting from the second S in the word PERSIST and taking every second letter gives you SPRITES

And that's about it. There aren't many other interesting examples, other than very short words. I figured it wouldn't make a very challenging puzzle.

But then I saw something unexpected and fascinating: There are over 100 words that yield themselves again! And the longest of these words are 12 letters long, giving solvers plenty to work with in an enigmatic puzzle.

It is particularly delightful because you get the same word back again, but the letters are now in a different order! Only, you can't tell that they've been rearranged.

If you're a visual person, you might appreciate this graphical representation:


As Gareth pointed out, for each of the 12-letter words I listed, it happens that you need to take every seventh letter. But there's no reason why this needed to be so. It could just as easily have been every fifth letter, for example.

(But it wouldn't work to take every third or fourth or sixth letter. To say more gets us deep into modular arithmetic, congruence classes, relatively prime pairs...)

  • $\begingroup$ Is the title "modulus" a typo? $\endgroup$
    – cinico
    Jul 10, 2019 at 19:59
  • $\begingroup$ @cinico — Glad you noticed that! It is not a typo. It is deliberate. $\endgroup$
    – SlowMagic
    Jul 10, 2019 at 23:04
  • $\begingroup$ @SlowMagic are you sure it is a riddle?? $\endgroup$ Jul 11, 2019 at 10:43
  • $\begingroup$ @OmegaKrypton — Feel free to edit the tags to make it better, if you like. $\endgroup$
    – SlowMagic
    Jul 11, 2019 at 15:36

3 Answers 3


Each word can be

reproduced by taking its letters in a different sequence of the form $(a+bn \textrm{ mod } 12)$. So e.g., if you start at the beginning of ANDROCENTRIC and take steps of size 7, wrapping in the obvious way, you get the word ANDROCENTRIC again, and likewise for RESETTLEMENT; if you start at the other C in CONDESCENDED and take steps of size 7, you get CONDESCENDED again; same for EXTRATEXTUAL and (second S) SUBMERSIBLES.

So as it happens

the step is always 7 letters; prima facie it could have been 5 instead. (Or 11, but that would be kinda boring. Or 1, which would have been really boring.) Curiously, there seem to be no examples with step 5, at least in the wordlist I just checked, but there are quite a few with step 7. More examples: BATHYSCAPHES, TARAMASALATA, INTERMIXTURE. The others in my wordlist are all either very obscure (CRANIOCLASIS, METASOMATISM, PALMOSPASMUS) or reduplications like HUGGERMUGGER and SHILLYSHALLY.

In case the above isn't enough to make it obvious,

the "modulus" in the title refers to the "mod 12" wraparound used.

This property is closely related to the one described in hexomino's answer, and for actual English words I believe it is equivalent. Proof:

First, suppose hexomino's property holds. If our word is of the form _A_B_C_A_B_C then starting at the first letter and moving by 7 every time leaves it unchanged. (Taking indices starting at 0, we have $7n=n$ when $n$ is even and $7n=n+6$ when $n$ is odd.) If our word is of the form A_B_C_A_B_C_ then starting at the seventh letter and moving by 7 every time leaves it unchanged.

On the other hand,

if "my" property holds so that letters $n$ and $a+7n$ are always the same, then (applying this twice) letter $n$ is the same as letter $a+7(a+7n)=8a+49n=8a+n$ mod 12. That is, letter $n$ is always the same as letter $n+8a$; equivalently, letter $n$ is always the same as letter $8a$. If $a$ isn't a multiple of 3, this is the same as saying that letter $n+4$ is always the same as letter $n$, implying a word of the form ABCDABCDABCD. In such a word, a step of 7 is the same as a step of -1, so ABCD must in fact be ABBA. So a word of the form ABBAABBAABBA satisfies "my" property but not hexomino's. I don't believe any actual English words are of this form :-). Next, suppose $a$ is an odd multiple of 3; that is, $a=3$ or $a=9$. In either case we get six pairs of letters that have to be identical. $a=3$ yields (0,3), (1,10), (2,5), (4,7), (6,9), (8,11) for a word of the form ABCADCEDFEBF. Similarly, $a=9$ yields (0,9), (1,4), (2,11), (3,6), (5,8), 7,10) for a word of the form ABCDBEDFEAFC. I don't believe any actual English words are of either of these forms, though it's not obvious and my belief is based on doing a computer search. Otherwise, $a$ is a multiple of 6 and now what we have is exactly hexomino's condition.

  • $\begingroup$ Nice spot and well-defined - also explains why that pattern could be spotted in @hexomino's answer :) +1 $\endgroup$
    – Stiv
    Jul 11, 2019 at 13:29
  • $\begingroup$ Yup! (And thanks for fixing my boneheaded mistake of typing CONDESCENDING when I meant CONDESCENDED.) $\endgroup$
    – Gareth McCaughan
    Jul 11, 2019 at 14:39
  • 1
    $\begingroup$ This is essentially equivalent to the property I identified, right (as in, a word which satisfies one, satisfies the other)? Although, I do understand the motivation behind your answer - to include all the letters in the definition. $\endgroup$
    – hexomino
    Jul 11, 2019 at 14:43
  • $\begingroup$ Yes, I think the two properties are in fact equivalent since 2*7=12, though when I wrote the answer above I hadn't realised that. $\endgroup$
    – Gareth McCaughan
    Jul 11, 2019 at 15:02
  • 1
    $\begingroup$ I'm now wondering whether any poets have seen fit to write 12-line poems (or 12-line stanzas in longer poems) with rhyme scheme ABCADCEDFEBF. $\endgroup$
    – Gareth McCaughan
    Jul 11, 2019 at 16:08

Not sure if it's the whole answer but, in each word,

It's possible to blank out either the odd or even numbered letters and get a repeating pattern (a group of three letters repeated).


ANDROCENTRIC $\rightarrow$ _N_R_C_N_R_C
CONDESCENDED $\rightarrow$ C_N_E_C_N_E_
EXTRATEXTUAL $\rightarrow$ E_T_A_E_T_A_
RESETTLEMENT $\rightarrow$ _E_E_T_E_E_T
SUBMERSIBLES $\rightarrow$ S_B_E_S_B_E_

  • $\begingroup$ Thank you for your solution and for your observations. You are correct that for the five specific example words I listed, your solution is equivalent. It makes me realize that in the statement of this puzzle I fell short in three ways: (1) I should have checked my example words more carefully for incidental patterns; (2) upon identifying an incidental pattern, I should have provided more example words to break the pattern; and (3) the comment I left for you was misleading. $\endgroup$
    – SlowMagic
    Jul 11, 2019 at 18:38
  • $\begingroup$ I should have said something more like, "You are correct, and the pattern you identified is equivalent in these cases. However, I am also looking for a description of a kind of self-symmetry which encompasses all the letters of the word." $\endgroup$
    – SlowMagic
    Jul 11, 2019 at 18:39

There is a 'self-symmetrical' pattern which exactly fits 4 of the 5 clues, and only fails on the 5th by a single letter:

The pattern of vowels and consonants is exactly the same in the first and second halves of the word


Using 'c' for a consonant and 'v' for a vowel:
ANDROCENTRIC --> ANDROC | ENTRIC == vcccvc | vcccvc
CONDESCENDED --> CONDES | CENDED == cvccvc | cvccvc
RESETTLEMENT --> RESETT | LEMENT == cvcvcc | cvcvcc
SUBMERSIBLES --> SUBMER | SIBLES == cvccvc | cvccvc

The only exception is EXTRATEXTUAL:

EXTRATEXTUAL --> EXTRAT | EXTUAL == vcccvc | vccvvc where R and U are the offending letters...


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