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.
 A: Assuming the Martians are using positional notation in their number-of-fingers base, $n$, we can write the quadratic equation as
$$5 x^2 - (\; 5 \cdot n + 0\cdot 1 \;) x + (\; 1\cdot n^2 + 2\cdot n + 5\cdot 1 \;) = 0$$
or, spotlighting $n$,
$$n^2 + n ( 2 - 5 x ) + 5 ( 1 + x^2 ) = 0$$
We know that both $5$ and $8$ satisfy the equation, so that (using base-10 notation!)
$$\begin{align}
x = 5: \qquad n^2 - 23 n + 130 &= 0 & (1) \\
x = 8: \qquad n^2 - 38 n + 325 &= 0 &(2)
\end{align}$$
Subtracting equation (2) from equation (1) gives
$$15 n - 195 = 0$$
Consequently, $n = 13$ ... which is, of course, "$10$" in Martian numerals.
A: 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.
A: Another approach. Given the fact that, the roots of the equation 
$$5x^2-50x+125=0\tag1$$ 
is $\alpha,\beta = 5,8$, we can write down
$$(x-8)(x-5)=0$$
$$x^2-13x+40=0$$
multiplying both sides by $5$, we have
$$5x^2-65x+200=0\tag2$$
Now $(1) = (2)$, is only that $(1)$ is in a non decimal base $b$ so we can safely write
$$5b^1 + 0b^0=65$$
$$b = \frac{65}{5} = 13$$
A: Let's go for a programming solution and let it be a non-"mathematical" language: PHP, called from a console.

php -r ' function b($x) {
             global $b;
             return (int) base_convert($x, $b, 10);
         }
         $b = 9;
         while (1) {
             $x = b(8);
             if (b(5) * $x * $x - b(50) * $x + b(125) == 0)
                 break;
             $b++;
         }
         echo "$b\n";
       '

Output: $13$.
Edit
As pointed out in the comments, PHP is not the most natural choice. So, here is a more natural choice: Python.
b = 9
while True:
    x = int("8", b)
    if int("5", b) * x**2 - int("50", b) * x + int("125", b) == 0:
        print b
        break
    b += 1

Just to show that PHP is not as weird, here is a (seemingly) less appropriate language: bash.

b=9
while true; do
  x=$(($b#8))
  if [ $(($b#5 * $x**2 - $b#50 * $x + $b#125)) -eq 0 ]; then break; fi
  let b=$b+1
done
echo $b

A: 13 fingers. Working in base b, the equation is a(x-5)(x-8) = 5x^2 - 50x + 125=0, so a=5. Then 5(5+8) = 50, so 5+8=10, so we must have b=13.
A: 13 fingers. Translate $5x^2-50x+125$ into base-$b$:
$$
5x^2-(5b)x+(b^2+2b+5)
$$
Since this has roots $x=5$ and $x=8$ we must have
$$
5x^2-(5b)x+(b^2+2b+5)=k(x-5)(x-8)=kx^2-13kx+40k
$$
so, equating coefficients,
$$
5=k,\quad 5b=13k,\quad b^2+2b+5=40k
$$
and so $b=13$. It's easy to check that the last equation is satisfied as well.
Perhaps the Martians had two six-fingered hands and a trunk. We won't know until xenoarchaeologists provide some evidence.
A: In an interview, you can impress the interviewer, by mentally calculating and determining the result. As other answers have mentioned, you need to express the equation in base different from 10 and then equate it with the roots of the equation. 


*

*From the options, its clear that the base is greater than 10. That means $5$ and $8$ are unit digits in some base b

*We know, for any quadratic equation $Ax^2+BX+C=0$, you can express the sum of the roots of the equation as $\alpha + \beta = -\frac{B}{A}$

*Any number in base b can be converted to base 10 by multiplying the digits with the nth power of the base, where n is the decimal position of the digit. So $50 = 5b^1+0b^0=5b$


Thus, if you know these concepts, you just need to solve the equation
$$-\frac{-50}{5}=-\frac{-5b}{5}=\alpha+\beta=5+8$$
$$b = 13$$
which is the base of the number system of Martians. 
Now correlating with human number system origin that base 10 is because we have 10 fingers, which would mean, Martians have 13 fingers\tentacles\..... 
A: Plugging in 8 gives even - even + odd = 0.  Clearly wrong, so 50 and 125 cannot be in base 10.  An even base won't change this, so the base must be odd (13 or 25).  25 is $5^2$ so the equation for 5 would be $5^3 - 5^4 + 5^5 + \dots$ in base 25.  This is clearly greater than 0, so it can't be 25.
Therefore if any answer is correct, it must be 13.
A: The correct answer is (a) 10.
There is no comment which number system the given answers refer to. As all other numbers refer to the Martian number system, we can safely assume the answers refer to the Martian number system as well.
A: Historically we kept using different number systems much later than we had 10 fingers. We have an innate ability of recognizing up to 3-5 items without any counting takes place. It is called subitizing. Some animals exhibit the same behavior too. So it's not about the fingers but about recognizing digits at one look. Base ten is great for that.
Fingers are extensions that are subject to functionality/resources trade-off just like branches or caterpillar feet. Especially a ridiculously complex move such as grasping requires precision hence lots of extra muscles and nerve complications. Think of all the animals that can and cannot grasp. You'll see a pattern. 
Without delving into any mathematics I would say object to the question that it is very unlikely to have that many fingers for grasp capable tentacle (it says the history is similar). 
So offer the game instead of the piece. 
A: Yet another way to find it, using the unexploited piece of data that we are proposed with possible solutions to avoid any non trivial mental calculus.


*

*$x=8$ being a solution rules out solution (a), base = 10.

*As $x=5$ is a root wathever the base b used by martians we will have: $$5.5.5-5.5.b+b^2+5.b+5=0$$  From wich we trivially get: $$25 b - 125 > b^2$$
Obviously 25 is already too large, it rules out solution (c) and it also rules out 40, solution (d).
We are left with solution (b), martian have 13 digits, which we should still check as a valid answer.
Of course other answers are better as they do not rely on fixed list of proposed possible solutions.
A: In response to Vedran Šego's Answer, here is the Python program which I suppose is the right way to solve this problem, if someone intents to solve it programatically
from sympy import solve
from sympy.abc import *
#Converts any decimal number to arbitrator base
def base(n):
    from itertools import count
    return sum(int(e) * b**i for e, i in zip(str(n)[::-1], count(0)))
#Given equation in arbiratory base
eqn = base(5)*x**2 - base(50)*x + base(125)
#solve the equation on b (base)
eqn = solve(eqn, b)
#now substitute one of the roots and to discard the trivial solution
#use the max function. 
print max(e.subs({'x':5}) for e in eqn)
13

A: (I was the OP. This is an excerpt from Adriano's answer with a tiny alteration. (Please provide credits only to him). This answer is just for the sake of my understanding and would not be selected as the best answer)
We can safely assume that $b≠10$ since $x=8$ would not be an answer in base-10.
Let the Martians have b fingers. So, their number system will be in base-b. 
So, the given equation  $5x^2 - 50x + 125 = 0$  is actually in base-b.
We should write it explicitly in base-b. 
$(5b^0)x^2 - ( 5b^1 + 0b^0)x + (1b^2 + 2b^1 + 5b^0) = 0$
Remember, our goal here is to find $b$.
They've said that $5$ and $8$ were the solutions given when it was calculated by Martians(wrt base-b).
So,   substitute $5$ in the equation(base-b) in the place of $x$ and solve it. You will get $b = 10, 13$
then substitute $8$ in the equation(base-b) in the place of $x$ and solve it. You will get $b =  25, 13$
Since both get the b value as $13$, it's the answer.
