Number of fingers of a Martian I have a question about what seems to be modular arithmetic, but I can't quite get the answer.
The problem goes along the lines of:
It is often said Earthlings use the decimal system because they have ten fingers. We see a Martian write down the equation:
$$
x^2 - 19x + 76 = 0
$$
When asked to write down the difference between the larger and smaller root, the Martian writes 9. How many fingers do Martians have?
Note: Martians write numbers between 0 and 9 exactly as we do.
 A: Hint:
Let $x_1$ and $x_2$ be the roots of the polynomial ($x_1 > x_2$) and let $b$ be the base martians use. Then:
$(x-x_1)(x-x_2) = x^2 - (x_1 + x_2) x + x_1\cdot x_2 = x^2 - 19_bx + 76_b$
And knowing that $x_1 - x_2 = 9$ then
$$\begin{cases}
x_1 - x_2 = 9 & \\
x_1 + x_2 = 1\cdot b + 9 & \\
x_1\cdot x_2 = 7\cdot b + 6 & \\
\end{cases}
$$
A: We have:
$x^2 - (b + 9)x + 7b + 6 = 0 \Rightarrow x_1 - x_2 = 2\sqrt{(\frac{b+9}{2})^2-7b - 6} = 9$
We can rearrange  little:
$(\frac{b+9}{2})^2-7b - 6 = \frac{81}{4}$
$(b+9)^2-28b - 24 = 81$
$b^2 + 18b + 81 -28b - 24 = 81$
$b^2 -10b - 24 = 0$
$b_{12} = 5 \pm \sqrt{5^2 + 24} = 5 \pm 7 \Rightarrow b = 12$
So martians can be assumed to have a dirty dozen of green fingers!
A: Call the roots $x_1$ and $x_2$. If $x_1-x_2=9$, we must have:
$$(x_1-x_2)^2=81_{10}$$
Or, in other words:
$$(x_1+x_2)^2-4x_1x_2=81_{10}$$
Using Viete, we have:
$$19_b^2-4\times76_b=81_{10}$$
Could that help?
A: Very interesting question.
The equation:
$x^2-19x+76=0$
In our world, the roots for this equation are
$$\frac{19 \pm \sqrt77}{2}$$
Difference : $\sqrt 77$
This same difference is 9 in his world, on a different base.
So, we can apply base conversion rules here.
$$\sqrt77_{10}=9_b$$
$$\log_{10}\sqrt77=\log_b9$$
$$\log_{10}\sqrt77=\frac{\log_{10}9}{\log_{10}b}$$
The answer, however comes out to be $10.25$ which is quite strange.
