The following is a list of 11 numbers in the Semur language¹ in a randomized order. Ten of them represent the first ten prime numbers. The remaining number is the sum of two other numbers on the list. Can you figure out the values of each number?

cê hnám lelka nâka
sén hnám lelka nâka
cê sén lelka nâka
sén phám lelka nâka
sén sén lelka nâka
cê qhê lelka nâka
cê phám lelka nâka
hnám qhê lelka nâka

¹Semur is a naturalistic constructed language made by myself. For all intents and purposes, you can treat it as if it was any other unknown human language.

  • 4
    $\begingroup$ Ooh, great puzzle! I'd love to see more linguistics puzzles around here. $\endgroup$
    – Deusovi
    Commented Feb 10, 2018 at 0:47
  • $\begingroup$ @Deusovi Thanks :) I’m currently working on one that is much more linguistically inclined than this one even, but it’s really hard to make sure it’s solvable, unabiguous and enjoyable. $\endgroup$ Commented Feb 10, 2018 at 10:51

1 Answer 1


Stream of consciousness (actual answer at the end); I have spoilered only the actual answer and one table before it because I don't think anyone glancing at this will find it tells them much unless they actually bother reading.

The first ten primes are 2,3,5,7,11,13,17,19,23,29.

It seems a safe bet that "ce", "pham", and "hnam" (my apologies, but I'm not going to bother with the diacritics) are 2,3,5 in some order. Why isn't 7 (which is after all quite small) a single word? Perhaps it's 10-3; or perhaps we're working in base 7 or smaller.

The multi-word numbers are all "X Y lelka naka". In at least some cases -- I'm going to guess all -- Y is itself a number. The simplest hypothesis is that "X Y lelka naka" means "X + bY" where b is the base the numbers are represented in.

Our candidate single-digit numbers, then, are: ce, hnam, sen, pham, qhe. So we're in base 5 at least. I claim we are in fact in base 6 or higher, because otherwise 29 would be a 3-digit number and I bet its representation wouldn't fit the pattern here.

OK; so, base 6 or base 7. Our numbers X are always ce or sen, which strongly suggests base 6, with one of those being 1 and the other 5; since ce appears in the list and sen doesn't, I think sen=1 and ce=5.

Clearly hnam and pham are 2,3 in some order. The table below shows what we know so far, with the first column being values if hnam,pham = 2,3 and the second if hnam,pham = 3,2.

5 5 ce
17 23 ce hnam
13 19 sen hnam
3 2 pham
11 11 ce sen
19 13 sen pham
7 7 sen sen
2 3 hnam
? ? ce qhe
23 17 ce pham
? ? hnam qhe

[At this point I had a brief detour because I did some arithmetic wrong. I haven't reproduced that here.]

Ah, now, at this point qhe has to be 4. So ce qhe is 29 (= 2+23 as required) which leaves hnam qhe as the sum-of-two-primes: it's either 2+6*4=26 or 3+6*4=27. It can't be 27 (the sum of two of our primes is even unless one of them is 2, and 27-2=25 which isn't prime) so it must be 26 (= 3+23); so hnam=2. Our final list, therefore:

5 ce
17 ce hnam
13 sen hnam
3 pham
11 ce sen
19 sen pham
7 sen sen
2 hnam
29 ce qhe
23 ce pham
26 hnam qhe

  • $\begingroup$ Nicely done, and quick at that :) $\endgroup$ Commented Feb 9, 2018 at 21:30
  • $\begingroup$ It was fun. And I am a mathematician :-). $\endgroup$
    – Gareth McCaughan
    Commented Feb 9, 2018 at 21:40
  • $\begingroup$ I think you made a typo in the last one, 29 not 25 right $\endgroup$
    – as4s4hetic
    Commented Feb 9, 2018 at 21:48
  • $\begingroup$ @as4s4hetic good catch, didn’t even notice that. The right number is mentioned in the text though, so it’s definitely just a typo $\endgroup$ Commented Feb 9, 2018 at 21:51
  • $\begingroup$ Fixed -- thanks. (Strictly, it wasn't exactly a typo. That detour-due-to-bad-arithmetic I mentioned involved a value of 25 instead of 29 for that one, and I didn't fix the table up completely. But I promise I had got the reasoning right by the time I posted :-).) $\endgroup$
    – Gareth McCaughan
    Commented Feb 9, 2018 at 21:54

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