# 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.

• Did you know: Not all human languages use base 10. For example, there's a rare language somewhere that uses base 8; apparently, they were counting on the spaces between the fingers rather than counting on the fingers themselves. (Also, you might have heard that the Mayans had base 20 and the Babylonians had base 60. – Akiva Weinberger May 12 '15 at 12:07
• @columbus8myhw For completeness: en.wikipedia.org/wiki/Octal#By_Native_Americans – SubSevn May 12 '15 at 16:18

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

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

• I don't get it: we have that $\;x_1x_2=76\;$ . In base $\;b=6\;$ this is written $\;204\;$ , so why do you think it can be written as you wrote? Also, if $\;b=24\;$ , then $\;76=34_{24}\;$, again not the form you gave. Am I missing something here? – Timbuc May 12 '15 at 12:15
• @Timbuc Both $19$ and $76$ are already written in base $b$. – Darth Geek May 12 '15 at 12:18
• Exactly my point: $\;x_1x_2=76\;$ in decimal basis as we do, so why do you think that in basis $\;b\;$ that'd be $\;7b+6\;$ ? – Timbuc May 12 '15 at 12:20
• @Timbuc The Martians were the ones who wrote the $76$ in the first place. Doesn't that mean it was written in base $b$? – Akiva Weinberger May 12 '15 at 12:26
• Well, that's true and now I understand better the comment by Darth. Thank you both! I still can't see the solution, though...but now I think I know better what direction to take. – Timbuc May 12 '15 at 12:29

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!

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?

• But then how do I know b from the last equation? Thanks! – Isaacs May 12 '15 at 12:23
• @Isaacs Perhaps you should expand $19_b$ into $b+9$, and expand $76_b$ into $7b+6$. After expanding, I think you'll get a quadratic in $b$. – Akiva Weinberger May 12 '15 at 12:24
• @Timbuc The solution ends up being $<19$ anyway. – Akiva Weinberger May 12 '15 at 12:34
• @columbus8myhw I see that now thanks to your other comment: as the martian wrote $\;76\;$ , it must mean $\;7b+6\;$ in their base $\;b\;$ ...thanks. – Timbuc May 12 '15 at 12:43

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.