<s>[INST] What would have been our number system if humans had more than 10 fingers? Try to solve this puzzle. Try to solve this puzzle:

The first expedition to Mars found only the ruins of a civilization.
  From the artifacts and pictures, the explorers deduced that the
  creatures who produced this civilization were four-legged beings with
  a tentatcle that branched out at the end with a number of grasping
  "fingers". After much study, the explorers were able to translate
  Martian mathematics. They found the following equation:
  $$5x^2 - 50x + 125 = 0$$
  with the indicated solutions $x=5$ and $x=8$. The value $x=5$ seemed
  legitimate enough, but $x=8$ required some explanation. Then the explorers 
  reflected on the way in which Earth's number system developed, and found
  evidence that the Martian system had a similar history. How many fingers would 
  you say the Martians had?
$(a)\;10$
$(b)\;13$
$(c)\;40$
$(d)\;25$

P.S. This is not a home work. It's a question asked in an interview.
 [/INST] Many people believe that since humans have $10$ fingers, we use base $10$. Let's assume that the Martians have $b$ fingers and thus use a base $b$ numbering system, where $b \neq  10$ (note that we can't have $b=10$, since in base $10$, $x=8$ shouldn't be a solution). Then since the $50$ and $125$ in the equation are actually in base $b$, converting them to base $10$ yields $5b+0$ and $1b^2 + 2b + 5$, so we now have:
$$
5x^2-(5b)x + (b^2+2b+5)=0
$$
Since $x=5$ is a solution, substitution yields:
$$ \begin{align*}
5(5)^2-(5b)(5) + (b^2+2b+5) &= 0 \\
b^2-23b+130 &= 0 \\
(b-10)(b-13) &= 0 \\
b&=10,13
\end{align*} $$
Since we know that $b\neq10$, we conclude that the Martians must have $13$ fingers. Indeed, this makes sense, because if $50$ and $125$ are in base $13$, then converting them to base $10$ yields $5(13)=65$ and $1(13)^2+2(13)+5=200$, so our equation becomes:
$$ \begin{align*}
5x^2-65x+200 &= 0 \\
x^2-13x+40&= 0 \\
(x-5)(x-8)&= 0 \\
x&= 5,8 \\
\end{align*} $$
as desired.
</s>