Number of fingers of a Martian

Question

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.

Answer 1

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

$$\begin{cases} x_1 - x_2 = 9 & \\ x_1 + x_2 = 1\cdot b + 9 & \\ x_1\cdot x_2 = 7\cdot b + 6 & \\ \end{cases} $$

Answer 2

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?

Answer 3

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.

Answer 4

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!