How many fingers do martians have? Text of problem: It is supposed that we use base 10 as our number system because we have ten fingers.
A martian, after seeing the equation
$x^2-16x+41=0$
writes the difference of the roots as $10$.
End
How many fingers do martians have ?
Note: For numbers between $0$ and $6$, Martians' writing is the same as ours.
I have absolutely no idea how to solve that.
 A: Hint: Suppose martians have $n$ fingers. Then they will interpret $16$ as $n+6$ and $41$ as $4n+1,$ by the note. So, you need a value of $n$ that is greater than $6,$ such that the roots of $$x^2-(n+6)x+4n+1=0$$ have a difference of $n$. Apply the quadratic formula and take it from there, if you can.
A: If you look at the quadratic formula, you'll see that the difference between the roots is the square root of the discriminant, divided by the leading coefficient. If the martian's base is $\beta$, he will read your equation as
$$ x^2 - (\beta+6)x + (4\beta+1) = 0 $$
and his conclusion that the difference between the roots is $10_\beta$ amounts to asserting
$$ \frac{\sqrt{(\beta+6)^2 - 4\cdot(4\beta+1)}}1 = \beta+0 $$
This ought to give you enough information to solve for $\beta$.
A: Note that the sum of the roots is $A+B=16$ from the quadratic, and the difference is $A-B=10$. Add these to obtain $2A=26$ or $A=13$ and then $B=3$ all in martian. So in martian also $3\times 13=41$ and it is easy from there.
