# 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

$(a)\;10$

$(b)\;13$

$(c)\;40$

$(d)\;25$

P.S. This is not a home work. It’s a question asked in an interview.

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*}