10

Text of problem: It is supposed that we use base 10 as our number system because we have ten fingers.

A martian, after seeing the equation

$x^2-16x+41=0$

writes the difference of the roots as $10$. End

How many fingers do martians have ?

Note: For numbers between $0$ and $6$, Martians' writing is the same as ours.

I have absolutely no idea how to solve that.

JeffC
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Nameless
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    Easy. Martians have 10 fingers (in their base, naturally). – Yakk Sep 21 '15 at 13:44
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    How does the martian know what all those symbols mean unless the martian already learnt our number system and hence knows we use a decimal system? Did the martian never see a simple sequence such as $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$, yet still manage to comprehend quadratic equations written in our notation? That seems even less likely than a martian existing in the first place. I know this isn't the point of the question, but it is so completely absurd that the question stops making sense. – hvd Sep 21 '15 at 14:02
  • @Yakk ,surely martians have more than two fingers otherwise he couldnt write the difference.@hvd it appeared on an exam examination,i am trying to make sense too. – Nameless Sep 21 '15 at 14:08
  • @Jhon we are told they are in at least base 6 and 6 fingers by the "Note" – jon_darkstar Sep 21 '15 at 14:18
  • The problem for me was actually extrapolating that dectail from the note.In the beginning i've read it as that martians have 7 numbers which in the shape are equal to ours but do represent different quantity. The note should have been more precise in saying that at least for number base system <$7$ ,martian number system is the same as our . – Nameless Sep 21 '15 at 14:33
  • Related: http://math.stackexchange.com/questions/460729 – Rick Decker Sep 21 '15 at 14:34
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    I read this as "how many fingers do martinis have" – stannius Sep 22 '15 at 04:42

3 Answers3

16

Hint: Suppose martians have $n$ fingers. Then they will interpret $16$ as $n+6$ and $41$ as $4n+1,$ by the note. So, you need a value of $n$ that is greater than $6,$ such that the roots of $$x^2-(n+6)x+4n+1=0$$ have a difference of $n$. Apply the quadratic formula and take it from there, if you can.

Cameron Buie
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16

If you look at the quadratic formula, you'll see that the difference between the roots is the square root of the discriminant, divided by the leading coefficient. If the martian's base is $\beta$, he will read your equation as $$ x^2 - (\beta+6)x + (4\beta+1) = 0 $$ and his conclusion that the difference between the roots is $10_\beta$ amounts to asserting $$ \frac{\sqrt{(\beta+6)^2 - 4\cdot(4\beta+1)}}1 = \beta+0 $$ This ought to give you enough information to solve for $\beta$.

hmakholm left over Monica
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  • There's really no need to invoke the quadratic formula at that step, when you know the solution has to be $\beta$. – Glen O Sep 21 '15 at 13:57
  • @GlenO: Could you expand on that? Note that $\beta$ is not a root, but the distance between the two roots. The sum of the roots is $\beta+6$, but it's not clear to me how you would find their difference without appealing to something like the quadratic formula. – hmakholm left over Monica Sep 21 '15 at 14:23
  • Ah, my apologies on that part, as I had failed to notice the mention of "difference". That being said, it's still doable, as one can consider roots $a$ and $a+\beta$, giving two equations in two unknowns... and one of the two is very easily solved for (subtracting the two equations gives $2a\beta-6\beta=0$). – Glen O Sep 21 '15 at 14:30
16

Note that the sum of the roots is $A+B=16$ from the quadratic, and the difference is $A-B=10$. Add these to obtain $2A=26$ or $A=13$ and then $B=3$ all in martian. So in martian also $3\times 13=41$ and it is easy from there.

Mark Bennet
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