Actual Puzzle

After crash-landing your spaceship on an uncharted planet, you run across the following drawing:

$\begin{matrix} ⊙ & × & × & ⊓ & ⊓ & ⊙ & & \\ ⊙ & ∿ & × & ⊓ & × & ⊳ & & \\\hline ∿ & ⊓ & ⊓ & ⊳ & ⊙ & × & × & \\ \end{matrix}$

How many fingers do the inhabitants of the planet have?


  • The above is produced with MathJaX. Starting with a copy of its source may ease the technical part of writing an answer.

  • Provenance: I remember encountering a similar puzzle in a popular-science or riddle book (I don’t remember which). I wanted to pose it to somebody else, but could not find such a puzzle on the web. Thus I designed one on my own.

  • There is some overlap with this puzzle, but other aspects are crucially different.

  • $\begingroup$ Instead of F fingers on N hands, it could equally be N fingers on F hands(/appendages). Assuming they don't use positional notation. But actually that still builds in the assumption that all hands(/extremities) have the same number of fingers, F. All we know is F*N >= 4. The solution is not uniquely determined. $\endgroup$
    – smci
    Commented Jan 11, 2022 at 2:24
  • $\begingroup$ @smci Why F*N >= 4? $\endgroup$ Commented Jan 11, 2022 at 2:27
  • $\begingroup$ Because there are 5 symbols (one of which could represent '0'). Hence if we assume hands/appendages having the same number of fingers, then #digits = F*N >= 4. When we see what looks like a carry-1, we can conclude we know the max value of a digit, hence we know (#digits-1) (either 6 or 5, depending on our assumptions). $\endgroup$
    – smci
    Commented Jan 11, 2022 at 2:43
  • 1
    $\begingroup$ @smci: The beauty of “How many fingers do they have?” is that it is equally ambiguous in referring to single hands, both hands, or whatever. Of course, we could be very specific and technical here, but then we also have to do the same about some of the other educated guesses going into the answer, which would render the question less lateral, surprising, and satisfying. $\endgroup$
    – Wrzlprmft
    Commented Jan 11, 2022 at 6:45
  • 1
    $\begingroup$ Ok, forget the off-by-one mention, that part I'm wrong. But my point that the assumption is that the number of fingers corresponds one-to-one with the number of digits is suspect; there are different Finger-counting schemes. Finger binary is one example; finger ternary is another. $\endgroup$
    – smci
    Commented Jan 11, 2022 at 9:18

3 Answers 3


It seems likely that they have

Three 'fingers' on each of two 'hands' (or some other features providing six digits)


Replacing the glyphs with our familiar digits: \begin{matrix}3&1&1&2&2&3&\\3&0&1&2&1&4&\\\hline0&2&2&4&3&1&1\end{matrix} Then rewriting it with our usual positional numbering system: \begin{matrix}&3&2&2&1&1&3\\&4&1&2&1&0&3\\\hline1&1&3&4&2&2&0\end{matrix} Results in a simple base-6 addition problem with solution.

Edit for detailed reasoning:

Working with the assumption that the glyphs represent distinct digits in a basic addition problem, we can first reverse the order of the glyphs to use our familiar system of writing numbers. \begin{matrix}&⊙&⊓&⊓&×&×&⊙\\&⊳&×&⊓&×&∿&⊙\\\hline×&×&⊙&⊳&⊓&⊓&∿\\\end{matrix} Then note that the $×$ must be $1$ as it is the carry value. \begin{matrix}&⊙&⊓&⊓&1&1&⊙\\&⊳&1&⊓&1&∿&⊙\\\hline1&1&⊙&⊳&⊓&⊓&∿\\\end{matrix} Now we see that $⊓$ must be $1+1=2$ because there cannot be a carry-over from $1+∿$ unless $⊓=0$ \begin{matrix}&⊙&2&2&1&1&⊙\\&⊳&1&2&1&∿&⊙\\\hline1&1&⊙&⊳&2&2&∿\\\end{matrix} We can then determine that $2+2=⊳\implies ⊳=4$ and $2+1=⊙\implies ⊙=3$ as there are no carries involved. \begin{matrix}&3&2&2&1&1&3\\&4&1&2&1&∿&3\\\hline1&1&3&4&2&2&∿\\\end{matrix} Finally $1+∿=2$ must include a carry with $∿=0$ \begin{matrix}&3&2&2&1&1&3\\&4&1&2&1&0&3\\\hline1&1&3&4&2&2&0\\\end{matrix} With $3_b+3_b=10_b$, we must be working in base $b=6$. Confirming that the end result is valid in base 6 verifies the solution.


I just want to point out that

the number of fingers could be completely unrelated to the number of mathematical digits they use. Furthemore, even if they are related, then they could either have 6 fingers on each hand or 6 fingers in total (3 on each hand, assuming there are two hands). So the most accurate answer is "we don't know for sure, but most likely 6 or 12 fingers in total".

  • 2
    $\begingroup$ I just want to point out that I already pointed that out when I wrote 'or some other features providing ## digits' :o) $\endgroup$ Commented Jan 10, 2022 at 22:58
  • $\begingroup$ Of course, the puzzle expects you to make some educated guesses. We also cannot know for sure that what is written there is what it is and not just some graffiti exclaiming “Wrzlprmft for overmind”. However, if we strip this lateral part from the question, it would be much less surprising and satisfying. $\endgroup$
    – Wrzlprmft
    Commented Jan 11, 2022 at 6:48

Answer (spoiler, I was wrong):

I'd say they have 5 fingers on each hand.


There are 5 distinct glyphs, ⊙, ×, ⊓, ∿, ⊳. If we wildly extrapolate from human counting methods, there would be one number, and therefore one glyph, for each finger on each hand. They could also have 10 fingers and the puzzle only uses 5 of the values.

The aliens could also do something weird, like order their digits from the middle to the outsides, alternating left and right. This could be a haiku with zero mathematical intent. Etc. But I assumed on looking at it that it was intended to be a basic addition problem with glyph substitution.

Getting the correct answer:

Cheating a bit, I looked at Daniel's answer. He decided to try solving in base 6. I'm not that smart, so I wrote a program to brute force everything to base 12 (anything higher takes too much time).

It assumes addition (multiplication would probably add a lot more digits to the product, subtraction and division couldn't add any unless one glyph is a negative sign or decimal point). It tests both forward (treating the leftmost digits as most significant) and reverse (treating leftmost digits as least significant). The minimum possible base is 5, since there are 5 glyphs.

The only valid answer is if we reverse the digits left-to-right, and use base 6. Precisely the answer Daniel gave in a much more clever manner. I suspect that if base 12 has no solutions, higher bases won't either, because of something related to carrying in the left column, but I don't think I can prove that mathematically.

Screen capture of my solver that says all the other answers have no valid solution up to base 12.


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