# 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.

• For an interview?! What company was that for? Aperture Science?? – zerosofthezeta Aug 6 '13 at 1:17
• @VISHNUVIVEK Just for completeness, every number system if base 10 from the user's viewpoint. cowbirdsinlove.com/43 – Snakes and Coffee Aug 6 '13 at 3:47
• Ask them how many fingers these folk had: en.wikipedia.org/wiki/Babylonian_mathematics – Rob P. Aug 6 '13 at 8:25
• This question appears to be off-topic because it is about alien civilisations. – Mikhail Katz Aug 6 '13 at 11:16
• Thanks @UncleZeiv, pity you only get 2 rep for the edit while lazy OP got ~93 – Tobias Kienzler Aug 6 '13 at 12:58

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.

• In base $10$, we can rewrite a number such as $321$ as $300 + 20 + 1$ or as $3(10)^2 + 2(10) + 1$. The digits $3$, $2$, and $1$ are the hundreds place, the tens place, and the ones place, respectively. If $321$ was actually in base $4$ (for example), then these digits would actually be the sixteenths place, the fourths place, and the ones place so that converting back to base $10$ would yield: $$3(4)^2 + 2(4) + 1 = 48 + 8 + 1 = 57$$ – Adriano Aug 6 '13 at 1:49
• Is it just a coincidence, then, that 5 + 8 = 13? – Throsby Aug 6 '13 at 4:23
• @AndréCaldas We can safely assume that $b \neq 10$ since $x = 8$ would not be an answer in base 10. We could also have known that $b > 8$ from the indicated solutions, but as it turns out we don't need this piece of information. – user2307487 Aug 6 '13 at 13:14
• The next interesting question is: what we can we conclude from this information about Martians? The first natural conclusion is that Martians cannot be symmetrical about any number of axes other than thirteen- since 13 is prime - unless, that is, they have a single fingered appendage, which would have to be located along the axis of symmetry. – Jack Aidley Aug 6 '13 at 15:20
• @Throsby: It is not exactly a coincidence, but it shows that the people designing the problem were not trying very hard. – Ross Presser Aug 6 '13 at 20:18

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.

• Hi! I couldn't understand the second line. How did you form the equation? – Vishnu Vivek Aug 6 '13 at 1:21
• @VISHNUVIVEK. In base-$b$ representation, the number expressed as $a_na_{n-1}\dots a_1a_0$ is interpreted to have the value $a_nb^n +a_{n-1}b^{n-1}+\cdots+a_1b+a_0$. For example, in base-10 the number 346 is interpreted as $3\cdot10^2+4\cdot 10^1+6\cdot 10^0$. – Rick Decker Aug 6 '13 at 1:39

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\.....

The correct answer is (a) 10.

There is no comment which number system the given answers refer to. As all other numbers refer to the Martian number system, we can safely assume the answers refer to the Martian number system as well.

• Heh, clever one. – Matsemann Aug 7 '13 at 7:59
• Reminds me of this. – Pece Aug 7 '13 at 14:29
• AH! Took me forever, until I read this answer, to understand what the question was asking. If it's left in base martian, has it really been translated? – user18862 Aug 8 '13 at 0:04

Assuming the Martians are using positional notation in their number-of-fingers base, $n$, we can write the quadratic equation as $$5 x^2 - (\; 5 \cdot n + 0\cdot 1 \;) x + (\; 1\cdot n^2 + 2\cdot n + 5\cdot 1 \;) = 0$$ or, spotlighting $n$, $$n^2 + n ( 2 - 5 x ) + 5 ( 1 + x^2 ) = 0$$

We know that both $5$ and $8$ satisfy the equation, so that (using base-10 notation!) \begin{align} x = 5: \qquad n^2 - 23 n + 130 &= 0 & (1) \\ x = 8: \qquad n^2 - 38 n + 325 &= 0 &(2) \end{align}

Subtracting equation (2) from equation (1) gives

$$15 n - 195 = 0$$

Consequently, $n = 13$ ... which is, of course, "$10$" in Martian numerals.

• Hi! how did you form the first equation? – Vishnu Vivek Aug 6 '13 at 1:26
• @VISHNUVIVEK: I formed the first equation by interpreting "$50$" and "$125$" in "base-$n$". For instance, in base-ten, "$50$" means "fifty": that is, "$5$ tens and no ones"; and "$125$" means "one-hundred twenty-five": ie, "1 ten-squared, 2 tens, and 5 ones". To translate to Martian, I simply replace the "ten"s in these expressions with the unknown "$n$", since Martian numerals are presumably based(!) on powers of $n$ rather than powers of ten. – Blue Aug 6 '13 at 1:32

Another approach. Given the fact that, the roots of the equation $$5x^2-50x+125=0\tag1$$ is $\alpha,\beta = 5,8$, we can write down $$(x-8)(x-5)=0$$ $$x^2-13x+40=0$$ multiplying both sides by $5$, we have $$5x^2-65x+200=0\tag2$$ Now $(1) = (2)$, is only that $(1)$ is in a non decimal base $b$ so we can safely write $$5b^1 + 0b^0=65$$ $$b = \frac{65}{5} = 13$$

• This was my approach as well. The only polynomial with roots $5$ and $8$ and leading coefficient $5$ is clearly, in Earthling notation, $5(x-5)(x-8)=5(x^2-13x+40)$, so this must be the polynomial the Martians had. From this it is pretty clear, be inspecting the middle coefficient, that their base is $13$. – Jeppe Stig Nielsen Aug 8 '13 at 10:42

Let's go for a programming solution and let it be a non-"mathematical" language: PHP, called from a console.

php -r ' function b($x) { global$b;
return (int) base_convert($x,$b, 10);
}
$b = 9; while (1) {$x = b(8);
if (b(5) * $x *$x - b(50) * $x + b(125) == 0) break;$b++;
}
echo "$b\n"; '  Output:$13$. Edit As pointed out in the comments, PHP is not the most natural choice. So, here is a more natural choice: Python. b = 9 while True: x = int("8", b) if int("5", b) * x**2 - int("50", b) * x + int("125", b) == 0: print b break b += 1  Just to show that PHP is not as weird, here is a (seemingly) less appropriate language: bash. b=9 while true; do x=$(($b#8)) if [$(($b#5 *$x**2 - $b#50 *$x + $b#125)) -eq 0 ]; then break; fi let b=$b+1
done
echo $b  • Why would you ever use PhP to solve maths? – Ali Caglayan Aug 6 '13 at 9:50 • Wow. I think that's the first time I've actually seen wouldn't of in the wild, @DannyBirch. – TRiG Aug 6 '13 at 12:16 • For those who don't like PHP, I've also included Python (and bash, in case you're masohistic types :-P). – Vedran Šego Aug 6 '13 at 12:48 • @VedranŠego: Use while true for Bash. – nneonneo Aug 6 '13 at 13:53 • @Vedran. Well, yes, but then, if you need a base > 36, it will fail anyway. I would be better to use a custom function for base conversion. – Jean-Claude Arbaut Aug 6 '13 at 23:50 13 fingers. Working in base b, the equation is a(x-5)(x-8) = 5x^2 - 50x + 125=0, so a=5. Then 5(5+8) = 50, so 5+8=10, so we must have b=13. • And you can check that$5\times 5 \times 8 = 13^2 + 2 \times 13 +5$– Henry Aug 6 '13 at 6:47 Plugging in 8 gives even - even + odd = 0. Clearly wrong, so 50 and 125 cannot be in base 10. An even base won't change this, so the base must be odd (13 or 25). 25 is$5^2$so the equation for 5 would be$5^3 - 5^4 + 5^5 + \dots$in base 25. This is clearly greater than 0, so it can't be 25. Therefore if any answer is correct, it must be 13. Historically we kept using different number systems much later than we had 10 fingers. We have an innate ability of recognizing up to 3-5 items without any counting takes place. It is called subitizing. Some animals exhibit the same behavior too. So it's not about the fingers but about recognizing digits at one look. Base ten is great for that. Fingers are extensions that are subject to functionality/resources trade-off just like branches or caterpillar feet. Especially a ridiculously complex move such as grasping requires precision hence lots of extra muscles and nerve complications. Think of all the animals that can and cannot grasp. You'll see a pattern. Without delving into any mathematics I would say object to the question that it is very unlikely to have that many fingers for grasp capable tentacle (it says the history is similar). So offer the game instead of the piece. • According to Richard Adams, rabbits can count to four, a "fact" that could help ease students into the study of non-decimal bases ... if kids ever read that book anymore. :/ – Blue Aug 6 '13 at 8:50 Yet another way to find it, using the unexploited piece of data that we are proposed with possible solutions to avoid any non trivial mental calculus. •$x=8$being a solution rules out solution (a), base = 10. • As$x=5$is a root wathever the base b used by martians we will have: $$5.5.5-5.5.b+b^2+5.b+5=0$$ From wich we trivially get: $$25 b - 125 > b^2$$ Obviously 25 is already too large, it rules out solution (c) and it also rules out 40, solution (d). We are left with solution (b), martian have 13 digits, which we should still check as a valid answer. Of course other answers are better as they do not rely on fixed list of proposed possible solutions. In response to Vedran Šego's Answer, here is the Python program which I suppose is the right way to solve this problem, if someone intents to solve it programatically from sympy import solve from sympy.abc import * #Converts any decimal number to arbitrator base def base(n): from itertools import count return sum(int(e) * b**i for e, i in zip(str(n)[::-1], count(0))) #Given equation in arbiratory base eqn = base(5)*x**2 - base(50)*x + base(125) #solve the equation on b (base) eqn = solve(eqn, b) #now substitute one of the roots and to discard the trivial solution #use the max function. print max(e.subs({'x':5}) for e in eqn) 13  (I was the OP. This is an excerpt from Adriano's answer with a tiny alteration. (Please provide credits only to him). This answer is just for the sake of my understanding and would not be selected as the best answer) We can safely assume that$b≠10$since$x=8$would not be an answer in base-10. Let the Martians have b fingers. So, their number system will be in base-b. So, the given equation$5x^2 - 50x + 125 = 0$is actually in base-b. We should write it explicitly in base-b.$(5b^0)x^2 - ( 5b^1 + 0b^0)x + (1b^2 + 2b^1 + 5b^0) = 0$Remember, our goal here is to find$b$. They've said that$5$and$8$were the solutions given when it was calculated by Martians(wrt base-b). So, substitute$5$in the equation(base-b) in the place of$x$and solve it. You will get$b = 10, 13$then substitute$8$in the equation(base-b) in the place of$x$and solve it. You will get$b = 25, 13$Since both get the b value as$13$, it's the answer. • The reason you get$10_{10}$as one of the solution of the equations when you substitute the root$5_{10}$in the equation is because,$5_{10}$is the solution of the equation$5_{10}x^{2_{10}}+50_{10}x+125_{10}=0$. In similar ways, you get$25_{10}$as one of the solution when you substitute the root$8_{10}$as it is the solution of the equation$5_{25}x^{2_{25}}+50_{25}x+125_{25}=0_{25}\$. Note, the subscript denotes the base of the number. – Abhijit Aug 7 '13 at 13:32