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.

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

1. From the options, its clear that the base is greater than 10. That means $5$ and $8$ are unit digits in some base b
2. We know, for any quadratic equation $Ax^2+BX+C=0$, you can express the sum of the roots of the equation as $\alpha + \beta = -\frac{B}{A}$
3. 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\.....