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On a distant island, there is a class of 200 students; 33 are boys and 101 are girls. In what numerical system are they calculating?

I know what numerical systems are pretty well, but this exercise semms weird to me. How can I even figure it out from the information given?

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  • $\begingroup$ Just try the various base. $\endgroup$
    – lulu
    Commented Jun 18 at 16:03
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    $\begingroup$ Well, in base $10$ we don't have $200$ equal to $33+101$ (assuming that everyone is either a boy or a girl, which I assume is the intent here). What about base $5$? or base $9$? Trial and error yields the solution rapidly. $\endgroup$
    – lulu
    Commented Jun 18 at 16:06
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    $\begingroup$ $3+1\equiv 0\pmod b$ $\endgroup$ Commented Jun 18 at 16:18
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    $\begingroup$ No. I use base $10$. What is base $4$? $\endgroup$
    – peterwhy
    Commented Jun 18 at 19:05
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    $\begingroup$ Could be plain old base 10 if there are a whole lot of nonbinary people in the class. $\endgroup$
    – Hearth
    Commented Jun 19 at 1:30

3 Answers 3

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The answer is base $4$, that is, assuming every student has to be a boy or a girl.

When we write something like $123$ in base $b$, what we really mean is $1\cdot b^2+2\cdot b+3\cdot1$.

So this question reduces to solving a quadratic equation, namely

$2b^2=(3b+3)+(b^2+1)$.

Rearranging, we find $b^2-3b-4=0$ and so we can find $b=4$ by factoring.

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    $\begingroup$ or $b=-1$, which is an extraneous solution $\endgroup$ Commented Jun 18 at 16:24
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    $\begingroup$ b=-1 is not necessarily extraneous. Perhaps the islanders use a weird version of base -1, where the rightmost digits are 1s, then -1, 1, -1, 1 etc. Thus "200" means there are 2 islanders, "33" means 0 boys, and "101" means 2 girls. $\endgroup$
    – Matt
    Commented Jun 19 at 2:21
  • $\begingroup$ Well, assuming boys or girls are the only options and the base is not negative, I did it like this: 200 > 134 so the base is < 10. Considering the last digits, 1+3 = 0 (i.e. 10) so the base is 4. $\endgroup$
    – Alex Gian
    Commented Jun 19 at 2:50
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If you set it up as an addition problem $$\ 101\\ \underline{+ 33} \\ \ 200$$ You can see that $1+3=10$, so the base must be $4$

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    $\begingroup$ Quick and neat! $\endgroup$
    – printf
    Commented Jun 19 at 3:08
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$200 - 101 = dd$ in base $d+1$.

$200 = 101 + 33$, so $200 - 101 = 33$.

Thus, $d = 3$ and the base is $4$.

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